Series of Real Numbers- I | Mathematics for Competitive Exams PDF Download

Introduction (Series of Real Numbers)

A finite series is given by the terms of a finite sequence, added together. For example, we could take the finite sequence
(2k + 1)¹⁰_k=1 
Then the corresponding example of a finite series would be given by all of these terms added together,
3 + 5 + 7 + .... + 21
We can write this sum more concisely using sigma notation. We write the capital Greek letter sigma, and then the rule for the kth term. Below the sigma we write ‘k = V . Above the sigma we write the value of k for the last term in the sum, which in this case is 10. So in this case we would have

and in this case the sum of the series is equal to 120
In the easy way, an infinite series is the sum of the terms of an infinite sequence. An example of an infinite sequence is

And then the series obtained from this sequence would be

With a sum going on forever. Once again we can use sigma notation to express this series. We write down the sigma sign and the rule for the k-th term. But now we put the symbol for infinity above the sigma, to show that we are adding up an infinite number of terms. In this case we would have

Key points: A finite series is given by all the terms of a finite sequence, added together.
A infinite series is given by all the terms of an infinite sequence, added together.
Definition
An expression of the form a₁ + a₂ + a₃ + ... + a_n + ... where each a_n is a real number, is called an infinite series of real numbers and is denoted by  a_n or ∑ a_n.a_n is called the nth term of the series ∑ a_n.

Example 1:

Example 2:  is a n infinite series- The sequence of partial sums looks like:

We saw above that this sequence converges to 2, so


Example 3: The number n!, read n factorial, is defined as the product of the first n positive integers; 
n! = 1 - 2 ..... n
<n!> is an important sequence. Its first few terms are
1, 2, 6, 24, 120, 720, ...
By convention, 0! is defined by 0! = 1.

• 1 + 1 + 1 + 1 + 1 + 1 + .... =  
• 1 + 1/2 + 1/3 + 1/4 + .... = 1/n
• 1/2 + 1/4 + 1/8 + ... = 1/2ⁿ
• 1 + 1/4 + 1/9 + 1/16 + ... = 1/n²
• 1- 1 + 1- 1 + 1 - 1 + ... = (-1)ⁿ
• 
• 
• 

Sequence of Partial Sums of Series

Recall : A sequence {a_n} is a function which assigns a real number a_n to each natural number n : 1, 2, 3, 4, 5......i.e., a sequence is an ordered list of real numbers : a₁, a₂, a₃, a₄, a₅, ...,

Example 1:  generates the sequence 
Definition: An infinite series  a_n is the sum of the numbers in the sequence {a_n}, i.e.,  a_n = a₁ + a₂ + a₃ + a₄ + a₅ + ... .

Example 2:  
We need a more precise definition of an infinite series  a_n. Begin by constructing a new sequence of partial sums by letting (This step by step by step process will be called the Sequence of Partial Sums Test for the infinite series a_n.)
S₁ = a₁,
S₂ = a₁ + a₂,
S₃ = a₁ + a₂ + a₃,
S₄ = a₁ + a₂ + a₃ + a₄,
S_n = a₁ + a₂ + a₃ + a₄ + ... + a_n.
We can now say that the value of the infinite series is precisely the value of the limit of its sequence of partial sums, i.e.,
 a_n = a₁ + a₂ + a₃ + a₄ + ....
=  (a₁ + a₂ + a₃ + a₄ + ... + a_n)
=  S_n.

Examples of Partial Sums

For the sequence 1, 1, 1, 1,..., we have 1 = N → ∞. Thus, is divergent.
For the sequence 1, -1, 1, -1,..., we have S₁ = 1, S₂ = 0, S₃ = 1, etc. In general , S_N = 1 for N odd and S_N = 0 for N even. Thus, (-1)ⁿ is oscillatory.

The Harmonic Series

If one computes the partial sums for  1/n one finds S₁ = 1, S₂ = 3/2 = 1.5, S₃ = 11/6 ≈ 1.87, S₁₀ ≈ 2/93, S₂₀ ≈ 3.40, S₁₀₀₀ ≈ 7.49, S_100,000 ≈ 12.09. If fact, S_N → ∞ So that  1/n diverges.
If one computes the partial sums for  1/n², then one obtains S₁ = 1, S₂ = 5/4 = 1.25, S₃ = 49/36 ≈ 1.36, S₁₀ ≈ 1.55, S₁₀₀ ≈ 1.63, S₁₀₀₀ ≈ 1.64.
In fact,1/n² = (2) = π²/6 ≈ 1.644934068
Definitions

• A series ∑ a_n is said to be convergent, if the sequence <s_n> of partial sums of ∑ a_n is convergent. If S_n = s, then S is called the sum of the series ∑ a_n. We write S = a_n.
• The series ∑ a_n is said to be divergent, if the sequence <S_n> of partial sums of ∑ a_n is divergent,.
• The series ∑ a_n is said to oscillate, if the sequence <S_n> of partial sums of ∑ a_n oscillates.

Example : (Geometric series) The series 1 + r + r² + r³ + ... (r > 0) converges if r < 1 and diverges if r > 1.

Convergence of a Geometric Series

a geometric series with ratio r diverges if Irl ≥ 1. If 0 < Irl <1, then the series converges to the sum

Proof: It is easy to see that the series oscillates if r = ±1. then s_n = a + ar +ar² + .... + ar^{n- 1}. multiplication by r yields
rS_n = ar + ar² +... +arⁿ.
Subtracting the second equation from the first produces S_n - rS_n = a - ar². Therefore. S_n (1 - r) = a(1 - rⁿ). and the partial sum is

If 0 < Irl < 1, it follows that rⁿ → 0 as n → ∞, and you obtain

which means that series converges and its sum is a/(1 - r). It is left to you to show that the series diverges if Irl > 1.
Example can be given, such as a_n = n is an unbounded sequence with no limit point and a_n = 1, if n is even; a_n = n, if n is odd is an unbounded sequence with a limit point 1.

Example 1: The series ∑ (- 1)^n-1 oscillates.

Example 2: Perform the following index shifts.
(a) Write ar^n-1 as a series that starts at n = 0.
(b) Write  as a series that starts at n = 3.
Solution:
(a) In this case we need to decrease the initial value by 1 and so the n in the term must increase by 1 as well.
ar^n-1 =  ar^(n+1)-1 = arⁿ
(b) For this problem we want to increase the initial value by 2 and so all the n's in the series terms must decrease by 2.

Note: The nature of a series is not affected by the addition, alteration or omission of a finite number of terms or by multiplication of all the terms by a fixed non-zero number.
(a) ∑ (u_n ± v_n) = s ± t
(b) ∑ ku_n = ks, k ∈ R

Example 3: Convergent and Divergent Geometric Series
(a) The geometric series


has a ratio of r = 1/2 with a = 3, Because 0 < Irl < 1, the series converges and its sum is

(b) The geometric series

has a ratio of r = 3/2. Because Irl ≥ 1. the series diverge.

Example 4: A Geometric Series for a Repeating Decimal
Use a geometric series to write  as the ratio of two integers.
Solution: For the repeating decimal , you can write


For this series, you have a = 8/10² and r = 1/10². So.

Theorem : Limit of nth Term of a Convergent Series
If a_n converges, then a_n = 0.
Proof: Assume that
 a_n =  S_n = L
Then, because S_n = S_n-1 + a_n and
it follows that

which implies that {a_n} converges to 0.
Theorem : nth Term Test for Divergence
If


Example 5: Using the nth Term Test for Divergence
(a) For the series 2ⁿ. we have

So, the limit of the nth term is not 0 and the series diverges.
(b) For the series  .

So, the limit of the nth term is not 0 and the series diverges.
(c) For the series  1/n, we have
 1/n = 0
Because the limit of the nth term is 0. The nth-Term Test for Divergence does not apply and you can draw no conclusion about convergence or divergence.

Tests of Convergence of Series

Some important comparison series

1. Geometric series
∑un = 1 + r + r² + r³ + ... + r^n-1 + rⁿ + ....

• is convergent, when    I r I < 1
• is divergent, when    r ≥ 1
• is oscillating finitely, when r = -1
• is oscillating infinitely, when r < -1

2. Hyperharmonic series (or p-series)

• is convergent when p > 1
• is divergent, when p ≤ 1

3. The Auxiliary Series
The series  is convergent if p > 1 and divergent if p ≤ 1.

Comparison Test

Test I. First Comparison Test
Let ∑u_n and ∑v_n be two positive term series such that
u_n ≤ k v_n   ∀ n ≥ m.        ...(1)
(k being a fixed positive number and m a fixed positive integers)
Then
(i) ∑v_n converges  ⇒  ∑u_n converges.
(ii) ∑u_n diverges   ⇒ ∑u_n diverges.
Proof: Let <S_n> and <T_n> be the sequence of partial sums of the series ∑u_n and ∑v_n respectively.
For n ≥ m, we have
S_n - S_m = (u₁ + u₂ + ... + u_m + u_m+1 + u_m+2 + ... + u_n) - (u₁ + u₂ + ... + u_m)
or S_n - S_m = u_m+1 + u_m+2 + ... + u_n.          ... (2)
Similarly,
T_n - T_m = v_m+1 + v_m+2 + ... + v_n.        .... (3)
From (1) and (2), we obtain
S_n - S_m ≤ k (v_m+1 + v_m+2 + ...v_n)
or S_n - S_m ≤ k (T_n - T_m) using (3)
or S_n ≤ k T_n +a,                  ... (4)
where a = S_m - k T_m is a fixed number.
(i) Suppose ∑ v_n converges.
The sequence <T_n> of partial sums of ∑v_n is bounded above i.e., there exists a positive real number t such that
T_n ≤ t   ∀ n.          ...(5)
From (4) and (5),
S_n ≤ kt + a   ∀   n.
Thus the sequence <S_n> of partial sums of ∑u_n is bounded above and so ∑u_n is convergent.
(ii) Suppose ∑u_n diverges.
Then  S_n = + ∞            .... (6)
From (4) and (6), we obtain

It follows that the sequence <T_n> of partial sums of the series ∑u_n diverges and so ∑v_n diverges.

Example 1: Test for convergence the series:
(i)  1/logn,
(ii)  1/n²logn
Solution: (i) We know log n < n for all n ≥ 2.
∴
Since ∑1/n diverges, so by First Comparison Test, 1/log n, diverges.
(ii) We know
Since ∑1/n converges, so by First Comparison Test, converges

Example 2: Test for the convergence of the series:

Solution: We that n ! ≥ 2^n-1 ∀ n ≥ 2.
∴
Now
 being a geometric series with common ratio 1/2 < 1 is convergent. Hence, by First Comparison Test, ∑1/n! is convergent.
Test II. Second Comparison Test
If ∑u_n and ∑v_n are two positive term series such that
     ....(1)
Then
(i) ∑v_n converges   ⇒ ∑u_n converges,
(ii) ∑u_{n diverges}   ⇒ ∑v_n diverges,
Proof: For n ≥ m, we have


Thus

∴ u_n ≤ k  v_n  ∀  n ≥ m,        ....(2)
where k = u_m/v_m is a fixed positive number.
(i) ∑v_n converges  ⇒  ∑u_n converges,
(ii) ∑u_n diverges    ⇒ ∑v_n diverges.
Test III. Limit Form Test
Let ∑u_n and ∑v_n be two positive term series such that
(I is finite and non-zero).
Then ∑u_n and ∑v_n converge or diverge together.
Proof: Since u_n/v_n > 0 for all n, so
Thus / > 0, as / ≠ 0
Let ε > 0 be some number such that / - ε > 0.
Since  u_n/v_n = /, so there exists a positive integer m such that

or / - ε < u_n/v_n < / + ε ∀ n ≥ m, where v_n > 0 ∀ n.
∴ (/ - ε) v_n < u_n < (/ + ε) v_n   ∀ n ≥ m.         ....(1)
From (1), we consider
u_n < (/ - ε) v_n ∀ n ≥ m.         ....(2)
(Here / + ε is a fixed positive number)
Applying First Comparison Test in (2), we obtain
   ... (A)
From (1), we consider
(/ - ε) v_n < u_n ∀ n ≥ m
or v_n < ku_n  ∀ n ≥ m,           ....(3)
where k = 1// - ε > 0 is a fixed number.
Applying First Comparison Test in (3), we obtain
  ...(b)
From (A) and (B), it follows that
∑u_n converges    ⇔         ∑v_n converges,
∑v_n diverges       ⇔          ∑v_n diverges.
Hence the two series ∑u_n and ∑v_n converge or diverge together.
Remark

• In order to apply the Limit form test to a given series ∑u_n, we have to select a series ∑v_n (which is usually a p-series) in which the n^th term of v_n behaves as u_n, for large values of n.

Example 3: Test each of the following series for convergence:
(i)
(ii)
Solution:
(i)
The n^th term of this series of

So ∑u_n and ∑v_n converge or diverge together (Limit From Test). Since  converges, so ∑u_n converges.
(ii)
The nth term is

Now  and finite.
So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n = ∑1/n diverges, so ∑u_n diverges.

Example 4: Test for convergence the series

Solution: We have

Let

Now

So ∑u_n and ∑v_n converge or diverge together.

Example 5: Test for converges the series

Solution: We have

Let

So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n = E 1/n² converges, so the given series ∑u_n also converges.

Example 6: Test for convergence the series whose n^th term is

Solution:

or


Let
so that ∑v_n converges.
Now

So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n converges, so the given series ∑u_n also converges.

Example 7: Test for the convergence of the series:

Solution: 

or

Let

Now


So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n converges, so the given series ∑u_n converges.

Example 8: Test for the convergence of the series whose n* term is
{(n³ + 1)^1/3 - n}.
Solution: Let


Now

So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n converges, so ∑u_n converges.

Example 9: Test for convergence the series

Solution: The n^th term of the given series is

Let

So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n = ∑1/n^5/2 converges, so the given series converges.

Example 10: Test the convergence of the series S .
Solution: We have

Then

Thus the two series ∑u_n and ∑v_n converge or diverge together. Since ∑v_n = ∑1/n diverges, so the given series ∑u_n diverges.

Example 11: Show that the series converges.
Solution: The n^th term of the given series is

∴ u_n/v_n = 1, so that ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n = ∑1/n² converges, so the given series ∑u_n also converges.

Example 12: Show that the series 1 + a + b + a² + b² + a³ + b³ + ..., 0 < a < b < 1 is convergent.
Solution: Since 0 < a < b < 1, aⁿ < bⁿ ∀ n ≥ 0.
Since ∑bⁿ is a geometric series with common ratio b, 0 < b < 1 ; ∑bⁿ is convergent. Hence, by First comparison test, the given series ∑u_n is convergent.

Ratio Test

Test IV. D'Alembert’s Ratio Test
Let ∑u_n be a positive terms series such that
  ... (1)
Then
(i) ∑u_n converges if / > 1,
(ii) ∑u_n diverges if /< 1.
Test fails if / = 1.
Proof. Case I. Let / > 1.
We can choose some ε > 0 such that / - ε > 1 or α > 1, α = / - ε.
Using (1), there exists a positive integer m₁ such that

Consider

or
    ...(2)
where
 , being a geometric series with common ratio 1/α < 1, is convergent.
Hence ∑u_n is convergent (by Second Comparison Test is applied in (2)).
Case II. Let / < 1.
We can choose another ε > 0 such that / + ε < 1
β < 1, β = / + ε.
Using (1), there exists a positive integer m₂ such that

or

Consider

or
       ....(3)
where ∑v_n = ∑1/βⁿ, being a geometric series with common ratio 1/β > 1, is divergent. Hence, by Second Comparison Test as applied in (3), ∑u_n is divergent.
Case III. We shall give examples of two series : one convergent and the other divergent but both satisfying

The series ∑u_n = ∑1/n is divergent, but

The series ∑u_n = ∑1/n² is convergent, but

Remark 1: Another equivalent form of Ratio Test is as follows:
If ∑u_n is a positive term series such that
         ...(1)
Then
(i) ∑u_n is convergent if / < 1.
(ii) ∑u_n is divergent if / > 1.
Proof: We have  m, where m = 1//.
Then ∑u_n converges if m > 1    ⇒    1// > 1 ⇒ / < 1,
and ∑u_n converges if m < 1    ⇒ 1// < 1 ⇒ / < 1.
Remark 2 : If ∑u_n is a positive term series such that
then ∑u_n is convergent.
Proof : It is given that the sequence  diverges to ∞.
So there exists a positive integer m such that

or

where ∑vn = ∑1/2ⁿ, being a geometric series with common ratio 1/2 <1, is convergent. Hence, by Second Comparison Test, ∑u_n is convergent.
Remark 3: If ∑u_n is a positive term series such that
then ∑u_n is convergent.
Proof : For ε = 1/2, there exists a positive integer m such that

The result now follows by Remark 2.

Example 1: Test for the convergence of the series:

Solution: Let

By applying ratio test -




Example 2: Test for convergence the series

Solution: We have

∴
By Ratio Test, the given series diverges.

Example 3: Test for convergence the series
(i)
(ii)
Solution: 
(i)
∴
By Ratio Test, ∑u_n converges of 1/x > 1 i.e., x < 1 and ∑u_n diverges if 1/x < 1 i.e., x > 1. The test fails for x = 1.
For

∴ 
So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n = ∑1/n² converges, so ∑u_n converges (for x = 1).
Hence the given series converges for x ≤ 1 and diverges for x > 1.
(ii) We have

∴
By Ratio Test, the given series converges if 1/x > 1 i.e., x < 1 and diverges if 1/x < 1 i.e., x > 1. The test fails for x = 1.
For x = 1, u_n =
Let

Now


So ∑u_n and ∑v_n converge or diverge together. Since ∑v_n converges, therefore ∑u_n also converges (for x = 1).
Hence the given series converges for x ≤ 1 and diverges for x > 1.

Example 4: Test for the convergence of the series:
(i)
(ii)
(iii)
Solution:
(i)
∴
By Ratio Test, ∑u_n converges if 1/x² > 1 i.e., x² < 1 i.e., x < 1 and diverges if x² > 1 i.e., x > 1. The test fails if x = 1.
For x = 1,

Let

Then

Since ∑v_n = ∑1/√n diverges, so ∑u_n diverges (for x = 1).
Hence the given series converges if x < 1 and diverges if x ≥ 1.
(ii) We have

∴
By Ratio Test, ∑u_n converges if 1/x > 1 i.e., x < 1 and diverges if x > 1. The test fails if x = 1.
For x = 1,

Let

∴
Since ∑v_n = ∑1/n√n converges, so ∑u_n converges (for x = 1).
Hence the given series converges if x ≤ 1 and diverges if x > 1.
(iii) We have


By Ratio Test, ∑u_n converges if 1/x > 1 i.e., x < 1 and diverges if x > 1. The test fails if x = 1.
For x = 1,

Let
As done in part (ii), ∑u_n converges if x = 1.
Hence the given series converges if x ≤ 1 and diverges if x > 1.

Example 5: Test for convergence the series:
(i) for all Positive values of x.
(ii) 
Solution:
(i)
Now

∴
Hence, by Ratio Test, ∑u_n converges.
(ii) We have
∴
By Ratio test, ∑u_n converges if 1/x > i.e., x < 1 and diverges if 1/x < 1 i.e., x > 1. For x = 1, we have

∴
So the given series diverges for x = 1. Hence the given series converges if x < 1 and diverges if x ≥ 1.

Example 6: Test for convergence the series:

for all positive values of x.
Solution: Ignoring the first term of the series, we have

∴
By Ratio Test, ∑u_n converges if 1/x > 1 i.e., x < 1 and ∑u_n diverges if x > 1. Ratio Test fails if x = 1.
For x = 1,

Clearly,

Since ∑1/n² converges, so by First Comparison Test, converges.
Hence ∑u_n converges for x ≤ 1 and diverges for x > 1.

Example 7: Test for convergence the series with n^th term:
(i)
(ii)
Solution: 
(i)
∴

∴  By Ratio Test, the given series converges if 1/x > 1 i.e., x < 1 and diverges if x > 1. The test fails if x = 1.
For

Let

So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n diverges, so ∑u_n diverges (for x = 1).
Hence the given series converges if x < 1 and diverges if x ≥ 1.
(ii) 
∴
By Ratio Test, the given series converges if 1/x > i.e., x < 1 and diverges if 1/x < 1 i.e., x > 1. The test fails if x = 1.
For x = 1,

Let

∴
So ∑u_n and ∑v_n converge or diverge together.
Since ∑v_n = ∑1/n diverges, so ∑u_n diverges (for x = 1).
Hence the given series converges if x < 1 and diverges if x ≥ 1.

Example 8: Test for the convergence of the series

Solution: We have

∴
 Hence ∑u_n is convergent, by Ratio Test.

Example 9: Test for the convergence of the series:

Solution: Ignoring the first term, we have

∴

By Ratio Test, ∑u_n converges if 1/x > 1 i.e., x < 1 and diverges if x > 1. Ratio Test fails if x = 1.
For x = 1,

i.e.u_n = 1 ≠ 0. So ∑u_n diverges for x = 1.
Hence ∑u_n converges if x < 1 and diverges if x ≥ 1.

Raabe’s Test

Let ∑u_n be a positive term series such that

Then (i) ∑u_n converges if / > 1, (ii) ∑u_n diverges if / < 1.
The test fails if / = 1.
Note: Raabe's test is applied when D'Alembert's ratio test fails
Proof. Case I. Let / > 1.
We can choose some ε > 0 such that / — ε > 1 or α > 1, α = / - ε.
From (1), there exists a positive integer m such that

or

Consider

or nu_n - nu_n+1 > α u_n+1   ∀ n ≥ m
or nu_n - (n + 1)u_{n + 1} > (α - 1) u_{n + 1} ∀ n ≥ m.
Putting n = m, m + 1, m + 2, .... n - 1 and adding, we get
mu_m - nu_n > (α - 1) (u_m+1 + u_m+2 + ... + u_n)
or (α - 1) (S_n - S_m) <mu_m - nu_m < mu_m,  (where S_n = u₁ + u₂ + ... + u_m + u_m+1 + ... + u_n)
Thus
S_n < S_m + m/α-1 u_m ∀ n.
where S_m + m/α - 1 u_m is a fixed positive number.
It follows that the sequence <S_n> is bounded above and so the given series is convergent.
Case II. Let / < 1.
We can choose another ε > 0 such that / + ε < 1.
From (1), there exists a positive integer m₁ such that

Since

or

or

Since ∑v_n = ∑1/n diverges, so by Second Comparison Test, ∑u_n diverges.
Case III. Consider the two series  and  where the first series is divergent and the second series is convergent but for both the series

Remark: Raabe’s Test is applied when D’Alembert’s ratio fails.
When Ratio test fails i.e., when  Raabe’s test may be applied.

Example 1: Test for convergence the series:
(i)
(ii)
Solution:
(i)

∴
By Ratio Test, ∑u_n converges if 1/x > 1 i.e., x < 1 and diverges if x > 1. The test fails if x = 1. We shall now apply Raabe’s Test.
For

∴
⇒
By Raabe’s Test, ∑u_n converges (for x = 1).
Hence the given series converges for x ≤ 1 and diverges for x > 1.
(ii) 

∴
By Ratio Test, ∑u_n converges if 1/x > 1 i.e., x < 1 and diverges if x > 1. The test fails if x = 1.
For x = 1,

now,

∴
By Raabe’s Test, ∑u_n diverges (for x = 1)
Hence the given series converges for x < 1 and diverges for x ≥ 1.

Example 2: Examine the convergence of the following series:
(i)
(ii)
Solution: (i) We have

∴
By Ratio test, ∑u_n converges if 1/x² > 1 i.e., x < 1 (as x > 0) and diverges if x > 1. The test fails for x = 1.
For x = 1,


∴
By Raabe’s test, ∑u_n converges (for x = 1).
Hence the given series converges for x ≤ 1 and diverges for x > 1.
(ii)

∴
By Ratio test, ∑u_n converges if 1/x² > 1 i.e., x < 1 (as x > 0) and diverges if x > 1. The test fails for x = 1.
For x = 1,

Now

∴
By Raabe’s Test, ∑u_n converges (for x = 1).
Hence the given series converges for x ≤ 1 and diverges for x > 1.
(ii)

∴
By Ratio test, ∑u_n converges if 1/x² > 1 i.e., x < 1 (as x > 0) and diverges if x = 1.
For x = 1,

Now

and

By Raabe’s Test, ∑u_n converges (for x = 1).
Hence the given series converges for x ≤ 1 and diverges for x > 1.

Example 3: Test for convergence the series:

Solution: On ignoring the first term, we have

∴
Thus Ratio test fails. We shall now apply Raabe’s test.

∴
Hence, the given series is divergent.

De Morgan and Bertrand’s Test

If ∑u_n be a series of positive terms; such that:

(i) if / > 1, ∑u_n is convergent
(ii) (ii) if / > 1, ∑u_n is divergent
Proof: In which follows, we shall compare ∑u_n with the auxiliary series

which is known to be convergent if p > 1 and divergent if p ≤ 1.
Now, we have



Thus
   ...(1)
Case (i). Let / > 1. Choose a number p such that / ≥ p > 1.
By comparison test of the second type the series ∑u_n will be convergent if there exists a positive integer m such that [ n ≥ m


 n ≥ p + terms containing n or log n in the denominator ...(2)
Taking limits on both sides of (2) as n → ∞ we get

⇒ / > p ⇒ I > 1 a s p > 1.
Thus, ∑u_n converges if / > 1.
Case (ii). Proceed as in case (i) yourself.

Logarithmic Ratio Test

If ∑u_n be a series of positive terms; such that:

(i) if / > 1, ∑u_n is convergent
(ii) if / > 1, ∑u_n is divergent
(iii) if / > 1, ∑u_n may converge or diverge. So test fails.
Proof: Let / > 1 and let us choose ε > 0 such that / - ε > 1. Let / - ε = λ so that / > 1.
Now

⇒ there exists a positive integer m such that

⇒
⇒ u_n/u_n+1 > e^λ/n, [ n ≥ m         ...(1)
We now that the sequence <(1 + 1/n)ⁿ> converges to e and hence
e ≥ (1 + 1/n)ⁿ, [ n ∈ N
⇒ e^λ/n ≥ (1 + 1/n)^λ         ... (2)
From (1) and (2), we have
    ...(3)
where v_n = 1/n^λ.
Since λ > 1, so ∑v_n converges. Then, using comparison test of second type it follows that the given series ∑u_n also converges.
(ii) Prove as in part (i).
Note 1. The above Logarithmic test is alternative to Raabe’s test and should be used when D'Alembert’s ratio test fails and when either e occurs in u_n/u_n+1 or n occurs as an exponent in u_n/u_n+1.
Note 2. When Raabe's test fails we may use De Morgan’s and Bertrand’s test given below. Again, when logarithmic test fails we may use second logarithmic ratio test.

Second Logarithmic Ratio Tests

Let ∑u_n be a series of positive terms such that

Then the series is
(i) convergent if / > 1
(ii) divergent if / < 1
(iii) no other decision is possible if / = 1.
Proof: In what follows, we shall compare ∑u_n with the auxiliary series

which is known (refer Art.) to be convergent if p > 1 and divergent if p ≤ 1. Now, we have



by binomial theorem
Thus
   .... (1)
Case (i). Let / > 1. Choose a number p such that / ≥ p > 1.
By comparison test of the second type, the series ∑u_n will be convergent if there exists a positive integer m such that [ n > m



[ ∵ log (1 + x) = x - x²/2 + x³/3 + .... ]


or / ≥ p. But p > 1, so / > 1.
Hence the given series ∑u_n converges if / > 1.
(ii) Proceed as in case (i) yourself.
Example 1: Examine the convergence of the following series:
(i)
(ii) 
(iii) 
(iv) 
Solution: (i) Here, we have

Here

Hence ratio test fails and we now apply Raabe’s test.

(ii) 
∴     ....(1)
and

Hence by ratio test, ∑u_n converges if 1/x² > 1, i.e., x² < 1, i.e., x < 1 (as x > 0) and diverges if 1/x² < 1, i.e., x² > 1, i.e., x > 1. The test fails if x = 1. In that case, from (1), we have

∴
and hence

Hence, by Raabe’s test, ∑u_n is convergent.
Thus the given series converges if x ≤ 1 and diverges if x > 1.
(iii) Omitting the first term of the given series as it will not change the nature of the series, we obtain

∴        ...(1)
Now

∴ By ratio test ∑u_n converges if 1/x² > 1, i.e., if x² < 1 and diverges if 1/x² < 1, i.e., if x² > 1. When 1/x² = 1, i.e., x² = 1, the ratio test fails. We shall now apply Raabe’s test.
From (1), for x² = 1,

∴
Hence, by Raabe’s test, ∑u_n converges for x² = 1.
Thus the given series converges if x² ≤ 1 and diverges if x² > 1.
(iv) Omitting the first term, the n^th term of the resulting series is

[Here the nth term of the A.P., 1, 5, 9, ... is 1 + (n - 1) × 4, i.e., 4n - 3 and the n^th term of the A.P., 2, 6, 10, ... is 2 + (n - 1) ×  4, i.e., 4n - 2]
and so

∴           ...(1)
So

∴ By ratio test ∑u_n converges if 1/x² > 1, i.e., if x² < 1 and diverges if 1/x² < 1, i.e., if x² > 1. When 1/x² = 1, i.e., x² = 1, the ratio test fails. We shall now apply Raabe’s test.
From (1), for x² = 1,

∴
Hence, by Raabe's test, ∑u_n converges.
Thus the given series converges if x² ≤ 1 and diverges if x² > 1.

Example 2: Test the convergence of the following series
(i)
(ii)
Solution: (i) Omitting the first term, we have

Now
           ....(1)
∴
By ratio test, ∑u_n converges if 1/ex > 1, i.e., if x < 1/e and diverges if 1/ex < 1, i.e., if x > 1/e. When 1/ex = 1, i.e., x = 1/e, the test fails. Since u_n/u_n+1 will involve e for x = 1/ e, we shall apply logarithmic test.
For x = 1/e, from (1)

∴
 n {log e - n log (1 + 1/n)}

Hence, by logarithmic test, ∑u_n diverges.
Thus the given series converges if x < 1/e and diverges if x ≥ 1/e.
(ii) Here

Now,



           ...(1)
Hence

as

By ratio test, ∑u_n converges if 1/ex > 1, i.e. if x < 1/e and diverges if 1/ex < 1, if x > 1/e When 1/ex = 1, i.e., x = 1/e, the test fails. We now apply logarithmic test.
For  x = 1/e, from (1),





⇒
⇒
Hence, by logarithmic test, ∑u_n diverges.
Thus the given series converges if x < 1/e and diverges if x ≥ 1/e.

Example 3: Test for converges the following series:
(i)
(ii)
Solution: (i) Here, we have

∴     ...(1)
Here

Hence the ratio test fails and we now apply Raabe’s test.
Using (1),
    ...(2)
∴
Hence the Raabe’s test fails and we now apply De-Morgan’s and Bertrand’s test. From (2),

∴
and so

Hence, by De-Morgan’s and Bertrand’s test, ∑u_n diverges.
(ii) Here

and

∴     ...(1)
Hence the ratio test fails and we now apply Raabe’s test. Using (1),
    ...(2)
∴
Hence, by Raabe’s test, ∑u_n converges if b - a > 1 and diverges if b - a < 1. When b - a = 1. Raabe’s test fails and we now apply De-Morgan's and Bertrand’s test.
For b - a = 1, from (2),
∴
∴
So

Hence, by De-Morgan’s and Bertrand's test, ∑u_n diverges. Thus the given series converges if b - a > 1 and diverges if b - a ≤ 1.

Example 4: Test for convergence the following series
(i)
(ii) x + x^{1 + 1/2} + x^{1 + 1/2 + 1/3} + x^{1 + 1/2 + 1/3 + 1/4} + ...
Solution: (i) Omitting the first term, we have

∴    ...(1)
Here  and so ratio test fails.
We shall now apply logarithmic test, Using (1), we get


= p(1/2n - 3/8n² + 7/24n³ + ....)
∴     ...(2)
So

Hence ∑u_n converges if p/2 > 1, i.e., if p > 2 and diverges if p/2 < 1, i.e., if p < 2 and the test fails if p = 2.
For p = 2, from (2).

∴
∴
Hence by second logarithmic ratio test ∑u_n diverges for p = 2.
Thus the given series converges if p > 1 and diverges if p ≤ 2.
(ii) Here, u_n = x^{1 + 1/2 + ... + 1/n},  u_n+1 = x^{1 + 1/2 + ... + 1/n + 1/(n +1)}
∴    ...(1)
Here ratio test fails and we now apply logarithmic test.
From (1),
   ...(2)
∴
Hence ∑u_n converges if log (1/x) > 1, i.e., if 1/x > e, i.e., if x < 1/e and ∑u_n diverges if log (1/x) < 1, i.e., if x > 1/e. When x = 1/e, the test fails and we shall now apply second logarithmic test
For x = 1/e, from (2)



Hence ∑u_n is divergent for x = 1/e.
Thus, the given series converges if x < 1/e and diverges if x ≥ 1/e

Gauss Test

Theorem: Let  a_n be a series with positive terms. Assuming that there exists a real number p, a real number r > 1 and a real bounded sequence {Bⁿ}^∞_n=1 such that for all n 

Then the series  converges if and only if p > 1.
Almost immediately we see that this test is somehow an improved version of Raabe’s test. The only difference is the number r, which in the former test was explicitly set to 2. So Gauss’ test is more general, allowing us to decide convergence of more series. Not surprisingly the proof is very similar.
Proof. If we start with p > 1, then for sufficiently large n we have (for negative terms of {B_n})

or (for positive terms of {B_n})

Since in both cases

by using Raabe’s test we can conclude that the series  a_n converges.
As for the other part of equivalence, we will assume that the series  a_n converges and p = 1. By using Bertrand’s test we get


With L’Hopital’s rule we get  (as n tends to infinity) and therefore the series  a_n diverges. A contradiction.
If the series a_n converges and p < 1 then (for sufficiently large n)

and according to Raabe’s test the series a_n diverges.
A contradiction. Thus p > 1.
Theorem: Let a_n be a series with positive terms. Assuming that there exist a real number p, a real number p, a real number r > 1 and a real bounded sequence {B_n} such that for all n.
    ... (1)
The seriesa_n diverges if and only if p ≤ 1.
Proof: Let a_n diverges if and only if p ≤ 1. Then for sufficiently large n.

and by Raabe’s test the series  a_n converges. A contradiction (hence p ≤ 1).
If p = 1 then by using Bertrand’s test (theorem)

With L’Hospital’s rule we get  (as n tends to infinity) and therefore the series a_n diverges.
If p < 1 then for sufficiently large n

We finish the proof with Raabe’s test.
An example. We want to determine the character of the series

With (x → ∞)

we get the final result

According to Raabe’s test, the series converges when a > 2 and diverges when a < 2 but we get no information when a = 2. We try Gauss’ test:
First, we try to find p

Second, we find A_n

Third, we try to find r > 1 such that B_n = A_nn^r is bounded

If we set r = 2 then

And because p = 1 the series   diverges.
Therefore, according to Gauss and Raabe’s test, the series  converges when a > 2 and diverges when a ≤ 2.

Kummer’s test

Here comes probably the most powerful test for convergence, since it applies to all series with positive terms.
Theorem: Let a_n be a series with positive terms. Then the series converges if and only if there exist a positive number A, positive numbers p_n and a number N ∈ N such that for all n > N

The series diverges if and only if there exist positive numbers p_n such that  and a number N ∈ N such that for all n > N
    ...(2)
Proof: First we prove the convergence. For the right-to-left implication, we adjust the equation (1)
p_na_n - p_n+1a_n+1 ≥ Aa_n+1
With qn = p_n/A we can write
q_na_n - q_n+1a_n+1 ≥ a_n+1
Since we know the left side of upper inequality, we can construct a sequence {B_n}^∞_n=1 such that
q_na_n - q_n+1a_n+1 = B_n+1a_n+1,   ∀n  B_n ≥ 1
The sequence {q_na_n}^∞_n=1 is positive and decreasing (follows from (3)).
Therefore it has a limit
0 ≤  a_nq_n < q₁a₁
Thus the series

converges. And because B_n ≥ 1 for all n

In words, the series  a_nB_n converges and creates an upper bound for the series  a_n, therefore the series a_n must converge as well. This is nothing more then just a first comparison criteria. We showed that finding numbers p_n is equally hard as finding a convergent series ∑b_n which creates an upper bound for the series ∑a_n in the first comparison test. For left-to-right implication, let p₁a₁ be a positive number. Now we assume the existence of positive monotonous sequence {B_n}^∞_n=1
 B_n = ∞           ...(4)
a_nB_n =P₁a₁ + a₁B₁
Now we shift the index n
a_n+1 B_n+1 = P₁a₁
We define the sequence {p_na_n}^∞_n=1 this way
p_n+1a_n+1 =p_na_n - a_n+1 B_n+1
where
p_na_n = p₁a₁ - a_k+1B_k+1 = 0
So, using (4) we have (for sufficiently large n)
p_na_n - p_n+1a_n+1 = a_n+1B_n+1 ≥ Aa_n+1 where A > 0
And finally

hence the number p_n are found.
Remark : In fact, the requirement (4) is not necessary as any positive and monotonous sequence {B_n} with  B_n = A ≥ 1 is totally sufficient (where A is an arbitrary constant).
For example, let ∑a_n be a convergent series. If we let B_n = 1 for all n and we construct numbers p_n using the terms from ∑a_n 

Kummer’s test will confirm the convergence of series ∑a_n (as would first comparison test). Now the divergence part.
Proof: For the right-to-left implication, we have ∑1/pⁿ = ∞ and from (2) we get

We can conclude the divergence of series a_n by using the second comparison test. To prove left-to-right implication (when a_n is divergent), we can assume, according to the theorem, the existence of positive and monotonous sequence B_n such that
B_n = 0      ...(5)
a_nB_n = ∞
We have

we get

It is not difficult to see, that ∑1/pⁿ = ∞ and p_n > 0 for all n, thus we found the numbers we were looking for.
Remark: Again, the requirement (5) is not necessary as any positive and non-increasing {B_n} B_n = A > 0 will do the trick (as it is the monotony we are interested in).
As in the previous remark, if ∑a_n is a divergent series and if we let B_n = 1 for all n then Kummer’s test will confirm the divergence.
Remark: To sum it up, Kummer’s test is very powerful because it really works for all the series with positive terms. On the other hand, using this test is equally difficult as using the first and second comparison test. The true strength of this test therefore lies in the numbers p_n. That is, the form of this test is a masterpiece, not its contents.
To demonstrate the power of Kummer’s test, we show that Raabe’s test and Bertrand’s test are in fact its corollaries. As for Raabe’s test, if we set p_n = n, we get

 compare with for convergence and

compare with for divergence.
We see, that what we can decide with Raabe’s test, we can also decide with Kummer’s test (with p_n = n) and vice versa, thus they are equivalent.
As for Kummer’s version of Bertrand’s test, we set p_n = n In n and put it into the left side of (1):

so
    .. (6)
Where

We go back to (1). With (6), if
∃A > 0, ∃N ∈ N, ∀n > N :
     ...(7)
then the series a_n is convergent.
Compare with Bertrand’s test:
If ∃A > 0, ∃N ∈ N,  ∀n > N:
     ...(8)
then the series a_n is convergent.
Since ε(n) can get arbitrarily small as n tends to infinity, we can hide it inside the positive constant A. Thus (7) and (8) are equivalent.
Divergence is a bit different and we will see that in this case, the tests are not equivalent. That is, Kummer’s test is slightly stronger. It is because now we have zero as a sharp border, while the constant A from the previous case was quite flexible.
With (2), (6) and p_n = n In n : if

then the series  is divergent. Compare with Bertrand’s test (see (5)) : If

then the series  a_n is divergent.
Now let's consider the series a_n, where a_n = 1/nInn. Using Bertrand’s test

(n + 1) In (n + 1) - n In n - In n - 1 = (n + 1) In (1 + 1/n) - 1 = ε(n) ≤ 0 → false Using Kummer’s test (with p_n = n In n)

0 ≤ 0 → true
There is infinite number of series that can be decided only with Kummer’s version of Bertrand’s test, but if we use limit, the tests are equivalent.

Root Test

Test IX. Cauchy’s nth Root Test
Let ∑u_n be a positive terms series such that
 (u_n)^1/n = I.     ...(1)
Then (i) ∑u_n converges if / < 1, (ii) ∑u_n diverges if / > 1. Test fails if / = 1.
Proof : Case I. Let / < 1.
We can choose some ε > 0 such that / + ε < 1 or α < 1, α = / + ε.
Using (1), there exists a positive integer m₁ such that
I (u_n)^1/n - I | < ε     ∀  n ≥ m₁.
or / - ε < (u_n)^1/n < / + ε        ∀ n ≥ m₁.
Consider (u_n)^1/n  < / + ε = α      ∀ n ≥ m₁.
un < αⁿ    ∀ n ≥ m₁.            ...(2)
Since ∑αⁿ = α + α² + .. being a geometric series with common ratio α < 1 is convergent, so by First Comparison Test as applied in (2), ∑u_n is convergent.
Case II. Let / > 1.
We can choose another ε > 0 such that / — ε > 1 or β > 1, β = / - ε.
Using (1), there exists a positive integer m₂ such that
I (u_n)^1/n - / I < ε    ∀ n ≥ m_2.
or / - ε < (u_n)^1/n < / + ε  ∀ n ≥ m₂.
Consider l - ε < (u_n)^1/n   ∀ n ≥ m₂.
or βⁿ < u_n         ∀ n ≥ m₂.        ...(3)
Since ∑βⁿ being a geometric series with common ratio β > 1 is divergent, so by First Comparison Test as applied in (3), ∑u_n is divergent.
Case III. We shall give examples of two series : one convergent and the other divergent, but both satisfying  (u_n)^1/n = 1.
The series ∑u_n = ∑1/n is divergent, but

The series ∑u_n = ∑1/n² is convergent, but