Sequences of Real Numbers- II | Mathematics for Competitive Exams PDF Download

The Algebra of Convergent Sequences

Theorem. If the sequence {a_n} converges to L and c ∈ R, then the sequence {ca_n} converges to cL; i.e.,  ca_n = c a_n.
Proof. Let’s assume that c ≠ 0, since the result is trivial if c = 0.
Let ε > 0. Since {a_n} converges to L, we know that there is an n ∈ N so that if n > N.

Thus, for n > N we then have that

which is what we needed to prove.
Theorem. If the sequence {a_n} converges to L and {b_n} converges to M, then the sequence {a_n + b_n} converges to L + M; i.e.,  (a_n + b_n) =  a_n +  b_n.
Proof. Let ε > 0. We need to find an N e R so that if n > N
l(a_n + b_n) - (L + M)| < ε.
Since {a_n} and {b_n} are convergent, for the given ε there are integers N₁, N₂ ∈ N so that
If n > N₁ then la_n - L| < ε/2 and if n > N₂ then lb_n - M| < ε/2 Thus, if n > max {N₁, N₂} then

Theorem. If the sequence {a_n} converges to L and {b_n} converges to M, then the sequence {a_n • b_n} converges to L • M; i.e., {a_n • b_n} =  a_n • b_n.
The trick with the inequalities here is to look at the inequality
|a_nb_n - LM| = |a_nb_n - a_nM + a_nM - LM|
< |a_nb_n - a_nMl + la_nM - LM|
= la_nllb_n - M| + |M||a_n - L|.
So for large values of n, lb_n - M| and |a_n - L| are small and |M| is constant. Now, by Theorem shows that la_nI is bounded, so that we will be able to show that la_nb_n - LMI is small.
Proof. Let ε > 0. Then there is a constant K > 0 such that la_nl ≤ K for all n. Since {b_n} is convergent, for the given ε there is an integer N₁ ∈ N so that
If n > N₁ then lb_n- M| < ε/2K.
Also, since {a_n} is convergent there is an integer N₂ ∈ N so that

Thus if N = max{N₁, N₂} then if n > N

Lemma, (limits preserve inequalities)

If the sequence {a_n} converges to L and {b_n} converges to M and if a_n < b_n for all n ≥ m, then L ≤ M.
Proof. Consider the sequence {c_n} defined by c_n := b_n - a_n.
The Sum Rule for Limits grants that c_n → M - L as n → ∞.
Furthermore, the assumption that a_n = y_n for all n ∈ N means that c_n ≥ 0 for all n ∈ N.
Applying the main result to the sequence {c_n} leads to the conclusion that M-L ≥ 0. That is, L ≤ M.

OR

Let if possible L > M.
Since lim a_n = L, therefore for ε = (L-M)/2, there exists n₁ such that Ia_n — L| < (L-M)/2 for all n ≥ n₁
i.e.. L - (L-M)/2 < a_n < L + (L-M)/2 for all n ≥ n₁
i.e. (L+M)/2 < a_n < (3L-M)/2 for all n ≥ n₁
Similarly, for b_n, there exists n₂ such that Ib_n - M | < (L-M)/2 for all n ≥ n₂
i.e. M - (L-M)/2 < b_n < M + (L-M)/2 for all n ≥ n₂
i.e. (-L+3M)/2 < b_n < (L+M)/2 for all n ≥ n₂
Let N = max (n₁, n₂)
Then, for all n ≥ N, b_n < (L+M)/2 < a_n
Let M = max (N, m)
For all n ≥ M, b_n < a_n, a contradiction
Hence L ≤ M.
Lemma. If the sequence {a_n} converges to L, if a_n ≠ 0 for all n ∈ N, and if L ≠ 0, then glb{la_nl In ∈ N} > 0.
Proof. Let ε = 1/2 ILI > 0. Since {a_n} converges to L, there is an N ∈ N so that if n > N then la_n - L| < |L|/2.
Now if n > N we must have that la_nl > I Ll/2. If not then the triangle inequality would imply

Now we set

Then clearly m > 0 and la_nl > m for all n ∈ N.
Theorem. If the sequence {a_n} converges to L, if a_n ≠ 0 for all n ∈ N, and if L ≠ 0, then the sequence {1/a_n} converges to 1/L; i.e.,  1/a_n = 1/a_n.
Proof. Let ε > 0. By Lemma there is a n, m > 0 such that la_nl > m for all n. Since {a_n} is convergent there is an integer N ∈ N so that if n > N IL — a_nl < ε • m|L|.
Then for n > N

Theorem. Suppose that the sequence {b_n} converges to M and if {a_n} converges to L. If b_n ≠ 0 for all n ∈ N, and if M ≠ 0, then the sequence {a_n/b_n} converges to L/M; i.e.,

Proof. We use two of the previous theorem to prove this. By Theorem {1/b_n} converges to 1/M, so

Theorem: Let {a_n} be a sequence converging to a then the sequence {la_nl} converges to lal:
Proof: This one, we break into three cases.
If a > 0, then (applying the definition of lim_n→∞ a_n = a with ε = a) there exists M_o so that a_n > 0
for n > M_o. In this case, we have la_nl = a_n for n > M_o and lal = a, and so lla_nl - |a|| = |a_n - a|. Since there is M₁ so that la_n - a| < ε for n > M₁ we have that lla_nl - |a|| < ε for n > M = max(M_o, M₁), and so lim_n→∞ la_nI = lal.
If a < 0, then (applying the definition of lim_n→∞ a_n = a with ε = lal) there exists M_o so that a_n < 0 for n > M_o. In this case, we have la_nl = - a_n for n > M_o and lal = - a, and so ||a_nl - |a|| = |-a_n + al = la_n - a|. Since there is M₁ so that la_n - a| < ε for n > M₁, we have that lla_nl - |a|| < ε for n > M = max(M_o, M₁), and so lim_n→∞ la_nl = lal.
If a = 0, then the definition of lim_n→∞ a_n = a becomes : for every ε > 0, there exists M so that la_n - 0| = |a_nI < ε for n > M. Since lla_nll = la_nI, we have that the definition of lim_n→∞ = 0 is satisfied without any further work.

OR

Let ε > 0 be given. Since lim a_n = a, therefore there exists n₁ such that la_n - a| < ε for all n ≥ n₁. Hence, I la_nI - lal I < la_n - a| < ε for all n ≥ n₁. Thus, lim la_nI = lal
Note: Converse of the theorem holds only for a = 0. Thus, lim a_n = 0 iff lim la_nl = 0. Converse does not hold in general. For example, if a_n = (-1)ⁿ, then lim la_nl = 1 but a_n does not even converge.

Example 1: Let p > 0 then lim_n→∞ 1/n^p = 0.
Let ε > 0 and let N = (1/ε)^1/p. Then n > N implies that n^p > 1/ε and hence ε > 1/n^p. Since 1/n^p > 0, this shows that n > N implies |1/n^p - 0| < ε.

Example 2: Let lal < 1, then lim_n→∞ aⁿ = 0.
Suppose a ≠ 0, because lim aⁿ = 0 is clear for a = 0. Since lal < 1, we can write
lal = 1/1+|b| where b > 0. By the binomial theorem

so

Now consider ε > 0 and let N = 1/εb.
Then if n > N, we have n > 1/εb and hence laⁿ - 0| < 1/nb < ε.

Example 3: (n^1/n) =1.
Let a_n = (n^1/n) - 1 and note that a_n ≥ 0 for all n.
By Theorem it is sufficient for us to show that lim_n→∞ a_n = 0.
Since 1 + a_n = n^1/n, we have n = (1 + a_n)ⁿ.
For n ≥ 2 the binomial theorem gives us

Thus,

Thus, we have shown that a_n <  for n ≥ 2.
Thus, lim a_n = 0.

Example 4:  (a^1/n) = 1 for a > 0.
Suppose a ≥ 1 then for n ≥ a we have 1 ≤ a^1/n ≤ n^1/n. Since lim n^1/n = 1, it easily follows that lim a^1/n = 1. Suppose that 0 < a < 1. Then 1/a > 1 so that lim (1/a)^1/n =1.
Thus,

Recall: For a sequence {a_n} we write lim a_n = +∞ provided for each M > 0 there is a number N such that n > N implies that a_n > M.
In this case we will say that {a_n} diverges to +∞.
We can make a similar definition for lim a_n = -∞.
Of course, we cannot use the previous theorem when dealing with infinite limits.
Theorem. Let {a_n} and {b_n} be sequence such that lim a_n = +∞ and lim b_n > 0. Then lim a_nb_n = +∞.
Proof. Let M > 0. Choose a real number m so that 0 < m < lim b_n. Whether lim b = +∞ or not, there exists N₁ so that if n > N₁ then b_n > m. Since lim a_n = +∞ there is an N₂ so that if n > N₂ then
a_n > M/m.
Setting N = max{N₁, N₂} means that for n > N, a_nb_n > M/m · m = M.
Theorem. For a sequence {a_n} of positive real numbers lim a_n = +∞ if and only if lim 1/a_n = 0.
Proof. Let {a_n} be a sequence of positive numbers. We need to show
If lim a_n = +∞ then lim 1/a_n = 0        (1)
and If lim 1/a_n = 0 then then lim a_n = +∞.   (2)
To prove (1) we will suppose that lim a_n = +∞. Let ε > 0 and let M = 1/ε. Since {a_n} diverges to +∞, there is an N so that if n > N then a_n > M = 1/ε. Therefore, if N > n then ε - 1/a_n > 0, so

Thus, this proves lim 1/a_n = 0.
To prove (2) suppose that lim 1/a_n = 0 and let M > 0. Let ε = 1/M. Then since ε > 0 there is an N so that if n > N then . Since a_n > 0 we then know that

and hence
if n > N then a_n > M.
This means that lim a_n = +∞ and 2 holds.
Theorem. All bounded monotone sequences converge.
Proof. Let {a_n} be a bounded monotonically increasing sequence and let S = {a_n ; n ∈ N}. Since the sequence is bounded, a_n < M for some real number M and for all n ∈ N. This means that the set S is bounded, and thus it has a least upper bound.
Let u = lub S and ε > 0. Since u = lub S and ε > 0, u - ε is not an upper bound for S. This means that there must be some N so that a_N > u - ε. Since {a_n} is monotonically increasing we have that for all n > N a_n ≥ a_N and hence for all n > N it follows that u - ε < a_n ≤ u. Thus, la_n - u| < ε for all n > N. Thus, lim a_n = u = lub S.
The proof for bounded monotonically decreasing sequences is the same with the greatest lower bound playing the role of the least upper bound.
Theorem. Let {a_n} be a sequence of real numbers.
(i) If {a_n} is an unbounded monotonically increasing sequence, then lim a_n = +∞.
(ii) If {a_n} is an unbounded monotonically decreasing sequence, then lim a_n = -∞.
Let {a_n} be a bounded sequence of real numbers. While it may converge or may not converge, the limiting behavior of {a_n} depends only on the “tails” of the sequence, or sets of the form {a_nI n > N}. This leads us to a concept that if a given sequence converges or diverges.
Let u_N = glb{a_n I n > N} = inf{a_n I n > N} and let v_N = lub {a_n I n > N} = sup{a_n I n > N}. We have seen that if lim a_n exists, then it must lie in the interval [u_N, v_N]. As N increases, the sets {a_n I n > N} get smaller, so we have
u₁ ≤ u₂ ≤ u₃ ≤ ... and v₁ ≥ v₂ ≥ v₃ ≥ ...
By the above theorem the limits u = lim_N→∞ u_N and v = lim_N→∞ v_N both exist and u ≤ v since u_N ≤ v_N for all N. If the limit exists, then u_N ≤ lim a_n ≤ v_N exists or not.
Theorem. A monotonic sequence of real numbers {x_n} converges if and only if it is bounded.
Proof: We already know that a convergent sequence is bounded. Hence we just need to show that a monotone and bounded sequence of real numbers is convergent.
Suppose {x_n} is a bounded increasing sequence.
Let S denote the non-empty bounded set {x_n : n ∈ N}
By the completeness axiom S has a least upper bound, and we let x = sup S.
We claim that lim x_n = x. Given any ε > 0, x - ε is not an upper bound for S.
Thus there exists an integer N such that x_N > x - ε.
Furthermore, since {x_n} is increasing and x is an upper bound for S we have
x - ε < x_N ≤ x_n ≤ x < x + ε
or equivalently
lx_n - x| < ε
for all n ≥ N. Hence lim x_n = n.
In the case when the sequence is decreasing, let x = inf S and proceed in a similar manner.
Lemma (On nested intervals). Let the sequence of the intervals {[a_n, b_n]} be such that [a_n+1, b_n+1] ⊂ [a_n, b_n] ∀n ∈ N (such intervals are called nested), and their lengths converge to zero, i.e., (b_n - a_n) = 0. Then there exists a unique point ξ that belongs to every interval, i.e., ξ ∈ [a_n, b_n] ∀n ∈ N, and ξ   a_n = sup{a_n} = b_n = inf{b_n}.
Proof. The sequence {a_n} and {b_n} are monotonic and bounded:
a₁ ≤ a_n ≤ a_n+1 ≤ b_n+1 ≤ b_n ≤ b₁ ∀n ∈ N.
By Theorem, they are convergent, and
a_n = sup {a_n} = a, b_n = inf {b_n} = b.
However, (b_n - a_n) = 0 implies that a = b = ξ: and clearly
a_n ≤ a = ξ = b ≤ b_n ∀ n ∈ N.
Next we show that there is no point other than ξ that belongs to all the intervals [a_n, b_n]. Assume the opposite: ∃η ∈ [a_n, b_n] ∀n and η ≠ ξ. Then by Theorem
b_n - a_n ≥ |ξ - n| >0 ∀n ⇒ (b_n - a_n ≥ |ξ - η| > 0.
This contradiction completes the proof of the lemma.

Example 5: Determine if the following sequences are monotonic and/or bounded.
(a) {-n²}^∞_{n =0}
(b) {(-1)ⁿ⁺¹}^∞_n=1
(c) {2/n²}^∞_n=5
Solution.
(a)  {-n²}^∞_{n = 0
This sequence is a decreasing sequence (and hence monotonic) because, - n}² _{> - (n + 1)}² for every n.
Also, since the sequence terms will be either zero or negative this sequence is bounded above. We can use any positive number or zero as the bound, M, however, it’s standard to choose the smallest possible bound if we can and it’s a nice number. So, we’ll choose M = 0 since, - n² ≤ 0 for every n
This sequence is not bounded below however since we can always get below any potential bound by taking n large enough. Therefore, while the sequence is bounded above it is not bounded below. we can also note that this sequence diverges (to -∞).
(b) {(-1)ⁿ⁺¹}^∞_{n = 1
The sequence terms in this sequence alternate between 1 and -1 and so the sequence is neither an increasing sequence or a decreasing sequence. Since the sequence is neither an increasing nor decreasing sequence it is not a monotonic sequence. It is an oscillatory sequence. The sequence is bounded however since it is bounded above by 1 and bounded below by -1.
Again, we can note that this sequence is not divergent, it is oscillatory.}
(c) {2/n²}^∞_{n = 5}
This sequence is a decreasing sequence (and hence monotonic) since,

The terms in this sequence are all positive and so it is bounded below by zero. Also, since the sequence is a decreasing sequence the first sequence term will be the largest and so we can see that the sequence will also be bounded above by 2/25. Therefore, this sequence is bounded.
We can also take a quick limit and note that this sequence converges and its limit is zero.

Example 6: Determine if the following sequences are monotonic and/or bounded.
(a) 
(b) 
Solution:
(a)
We'll start with the bounded part of this example first and then come back and deal with the increasing/decreasing.
First, n is positive and so the sequence terms are all positive. The sequence is therefore bounded below by zero. Likewise each sequence term is the quotient of a number divided by a larger number and so is guaranteed to be less than one. The sequence is then bounded above by one. So, this sequence is bounded.
Now let’s think about the monotonic question. First, we will often make the mistake of assuming that because the denominator is larger the quotient must be decreasing. This will not always be the case and in this case we would be wrong. This sequence is increasing.
To determine the increasing/decreasing nature of this sequence we will need to calculate I technique. First consider the following function and its derivative.


We can see that the first derivative is always positive and so from calculation we know that the function must then be an increasing function. Notice that,

Therefore because n < n + 1 and f(n) is increasing we can also say that,

In other words, the sequence must be increasing.
Note that now that we know the sequence is an increasing sequence we can get a better lower bound for the sequence. Since the sequence is increasing the first term in the sequence must be the smallest term and so since we are starting at n = 1 we could also use a lower bound of 1/2 for this sequence. It is important to remember that any number that is always less than or equal to all the sequence terms can be a lower bound.
(b)
This is a messy looking
Now, let’s move on to the increasing/decreasing part. As with the problem, many of us will look at the exponents in the numerator and denominator and determine based on that sequence terms must decrease.
This however, isn’t a decreasing sequence. Let’s take a look at the first few terms to see this.

The first 10 terms of this sequence are all increasing and so clearly the sequence can’t be a decreasing sequence. Recall that a sequence can only be decreasing if ALL the terms are decreasing.
Now, we can’t make another common mistake and assume that because the first few terms increase then whole sequence must also increase. If we did that we would also be mistaken as this is also not an increasing sequence.
This sequence is neither decreasing or increasing. The only sure way to see this is to do the Calculus approach to increasing/decreasing functions.
In this case we’ll need to the following function and its derivative.


This function will have the following three critical points,

Remember these are the only points where the function may change sign! Our sequence starts at n = 0 and so we can ignore the third one since it lies outside the values of n that we’re considering. By plugging in some test values of x we can quickly determine that the derivative is positive for 0 < x < ∜30000 ≈ 13.16 and so the function is increasing in this range. Likewise, we can see that the derivative is negative for x > ∜30000 ≈ 13.16 and so the function will be decreasing in this range.
So, our sequence will be increasing for 0 ≤ n ≤ 13 and decreasing for n ≥ 13. Therefore the function is not monotonic.
Finally, note that this sequence will also converge and has a limit of zero.

Example 7: Let x₁ = 6 and x_n+1 = 5 - 6/x_n for all n ∈ N then the sequence converges to.
Solution: x₁ = 6

Let x_k > 3 for some k ∈ N
then

We have x₁ > 3 and if we assume that x_k > 3 for some k ∈ N, then x_k+1 > 5 - 2 = 3. Hence by the principle of mathematical induction, x_n > 3 for all n ∈ N. Therefore (x_n) is bounded below. Again, x₂
= 4 < x₁ and if we assume that x_k+1 < x_k. for some k ∈ N, then X_K+2 - X_k+1 = 6 (1/x_k - 1/x_k+1) < 0 ⇒ x_k+2 < x_k+1. Hence by the principle of mathematical induction, x_k+l < x_n for all n ∈ N. Therefore (x_n) is decreasing. 
<x_n> is bounded and monotonic. Hence <x_n> is convergent. Let ℓ = X_n. Then x_n+1

⇒ ℓ = 5 - 6/ℓ (since x_n > 3 for all n ∈ N, ℓ ≠ 0) ⇒ (ℓ - 2)(ℓ - 3) = 0 ⇒ ℓ = 2 or ℓ = 3. But x_n > 3 for all n ∈ N, so ℓ ≥ 3. Therefore ℓ = 3.

Example 8: The series converges or diverges
Solution: The given series is a sum of two series. We will look at each of them separately and hope that the two results can be put together. First consider the series

This is a series that can be converted to a geometric series.

Since q = 1/5 satisfies Iql < 1, this series converges.
Now we will look at the series

When we see factorial, we immediately think of the Ratio test.

Since λ > 1, this series diverges.
Since the given series is the sum of a convergent and a divergent series, it is divergent.

Subsequence

Definition: Let <a_n> be any sequence. Let <n₁, n₂, ...., n_k ...) be a sequence of positive integers such that i > j ⇒ n_i > n_j Then the sequence (a_n1,a_n2,...., a_nk....}, written as (a_nk), is called a subsequence of (a_n).
Let {a_j}^∞_j=1 be a sequence. When we extract from this sequence only certain elements and drop the remaining ones we obtain a new sequences consisting of an infinite subset of the original sequence.
That sequence is called a subsequence and denoted by {a_jk}^∞_k=1
<a₁₀, a₁₁, a₁₂, ...) is a subsequence of <a_n>.
Theorem: If sequence bdd., then every subsequence is bdd and conversely.
Proof. Let {a_n} be any bdd. sequence
then ∃ m, M ∈ R such that
m ≤ a_n < M ∀ n ∈ N ...(1)
Let {a_nk} be any subsequence of {a_n}
⇒ m ≤ a_nk ≤ m (from (1)) ∀ n_k ∈ N
∴ {a_nk} is also bdd subsequence

Conversely

Let every subsequence of {a_n} be bdd.
As every sequence is a subsequence of itself.
∴ {a_n} is bdd.
Note: If we have to show {a_n} is unbounded.
Then show ∃ a subsequence of {a_n} which is unbounded.
Ex.

Consider


So, {a₂ⁿ} is unbounded
Hence, {a_n} is unbounded.

Observation

(i) A sequence has a constant subsequence iff ∃ a real no. which appears in the sequence infinite many times.
(ii) If range of a sequence is a finite set then there exist a real no. which appears in the sequence infinite many times.
Theorem: If a sequence <a_n> converges to I, then every subsequence of <a_n> converges to I.
Proof: Let <a_n> be a subsequence of <a_n>.
Since a_n → I, therefore, for any ε > 0, there exists a positive integer m such that
la_n - II < ε          ∀ n ≥ m.
In particular, |a_nk — I|< ε   ∀ n_k ≥ m.
Hence (a_nk} converges to I.
Remark: The converse of the above theorem may not be true.
Theorem: Every bounded sequence has a convergent subsequence.
Proof: Let <x_n> is bounded
Case 1: When the range of the sequence is finite.
If the range of <x_n> in finite then at least one element ‘a’ repeats infinitely because a sequence has infinite number of terms, if we choose this element ‘a’ again and again then the subsequence <a, a, ...> formed is convergent.
Case 2: When the range of the sequence is infinite
<x_n> is bounded and range is infinite bounded subset of set of real no. R we know that set of bounded real no. has at least one limit point. Therefore, let I is the limit point of <x_n>.
For each n ∈ N, ∃ v_n ∈ N s.t.
Let < x_V1, x_V2,...x_Vn,...> be a subsequence of <x_n>.
∴ for given ε > 0 ∃ n_o t ∈ N s.t. 1/n_o < ε
then

⇒ < x_v1, x_v2, ... x_vn, ... > is convergent.
Theorem: If the subsequences {a_2n-1} and {a_2n} of a sequence {a_n} converge to the same limit I, then the sequence {a_n} converges to I.
Proof. Let ε > 0 be given
{a_2n-1} converges to I 
⇒ For ε > 0, ∃ a positive integer m₁ such that      la_2n-1 - II < ε ∀ n ≥ m₁
{a_2n-1} converges to I 
⇒ For ε > 0, ∃ a positive integer m₂ such that      la_2n - II < ε ∀ n ≥ m₂
Let m = max. {m₁, m₂}, then la_2n-1 - Il < ε and    la_2n - II < ε ∀ n ≥ m
⇒ la_n — II < ε     ∀ n ≥ m
⇒ {a_n} converges to I.

Exercise. Take the sequence {(-1)^j}^∞_j=1. Extract every other member, starting with the first. Then do the same, starting with the second.
The sequence in question is
{(-1)^j}^∞_j=1 = {-1, 1, -1, 1, -1, 1, ...}
If we extract every second number, starting with the first, we get:
{- 1, - 1, - 1, - 1, ...}
This subsequence now converges to -1.
If we extract every second number, starting with the second, we get:
{1, 1, 1, 1, 1, ...}
This subsequence now converges to 1.

• Examples: The sequence  in R² is a subsequence of the sequence  (in this case, n_j := j²). The sequence 1, 1, 1, 1, ... is a subsequence of 1, 0, 1, 0, 1, ...

Exercise. Take the sequence {1/j}^∞_j=1. Extract three different subsequences of your choice and look at the convergence behavior of these subsequences.
The sequence in question is:
{1/j}^∞_j=1 = {1, 1/2, 1/3, 1/4, 1/5, 1/6,...}
which converges to zero. Now let us extract some subsequences:
Extracting the even terms yields the subsequence
{1/2, 1/4, 1/6, 1/8, 1/10, ...}
which converges to zero.
Extracting the odd terms yields the subsequence {1, 1/3, 1/5, 1/7, 1/9, ...} which converges to zero.
Extracting every third member yields the sequence {1, 1/4, 1/7, 1/10, 1/13, ...} which converges to zero.
Hence, all three subsequences converge to zero. This is an illustration of a general result: If a sequence converges to a limit L then every subsequence extracted from it will also converge to that limit L.

• On the other hand, it is possible for a subsequence to be convergent without the sequence as a whole being convergent. For example, the sequence 1, 0, 1, 0, 1, ... is not convergent, even though certain subsequences of it (such as 1, 1, 1, ... converges).

Cauchy Sequence

Definition. A sequence <a_n> of real numbers is said to be a Cauchy sequence if for every ε > 0 there exists a natural number N such that for all natural numbers n, m > N, the terms a_n, a_m satisfy la_n - a_ml < ε.
Definition. A sequence <a_n> is said to be a Cauchy sequence if for any ε > 0, there exists a positive integer m such that
la_n - a_ml < ε, whenever n ≥ m.
Note: It can be easily shown that the two definitions are equivalent. Most of the authors use the 1^st one as definition but in practice we’ll often use the latter one.
Theorem: Cauchy Sequence is bounded.
Proof: Let <a_n> be a Cauchy sequence.
For ε = 1, there exists a positive integer m such that
la_n - a_mI < 1,         whenever n ≥ m,
or a_m - 1 < a_n < a_m + 1,   whenever n ≥ m.
Let k = min {a₁, a₂, ...., a_m-1, a_m - 1},
K = max {a₁, a₂,...., a_m-1, a_m + 1}.
Then k ≤ a_n ≤ K for all n.
Hence <a_n> is bounded.
Theorem: If a subsequence of a Cauchy sequence converges to I, then the sequence also converges to I.
Proof: Let < x_n> be a cauchy sequence, then by definition for any ε > 0 ∃ n_o s.t.
lx_m- x_nl < ε/2    ∀ m, n > 0      ...(1)
Let < x_vn> be the subsequence of < x_n> and lim x_vn = I, then for I > 0 ∃ n_o ∈ N s.t.
|x_n — II < ε/2     ∀ v_n > n_o       ...(2)
Now |x_n - II = lx_nIl ≤ | x_Vn - x_n| + |x_Vn - I|
< ε/2 + ε/2     ∀ n > v_n [by (1) and (2)]
⇒ |x_n - II < ε   ∀n > v_n
⇒ lim x_n= I 
Theorem: A sequence X = <x_n> in R is convergent if and only if it’s Cauchy.
Proof: "⇒"
If x := lim X, then given ∈ > 0 there is a natural number K such that if n ≥ K then IX_n - x I < ∈/2. Thus, if n, m ≥ K, then we have lx_n - x_mI = I (x_n - x) + (x - x_m) I ≤ lx_n - x I + lx_m - x I < ∈/2 + ∈/2 = ∈-Since ∈ > 0 is arbitrary, it follows that <x_n> is a Cauchy sequence.
(Here we used the first definition)
“ ⇐ ” let X = (x_n) be a Cauchy sequence; we will show that X is convergent to some real number. First we observe that the sequence X is bounded. Therefore, by the Bolzano-Weierstrass Theorem, there is a subsequence X ’ = < x_nλ> of X that converges to some real number X*. We shall complete the proof by showing that X converges to X*.
Since X = (x_n) is a Cauchy sequence, given > 0 there is a natural number N such that if n, m N then lx_n -x_ml<.
Since the subsequence X’ = () converges to x*, there is a natural number K>N belonging to the set {n₁, n₂, ...} such that |x_k - x*| <.
Thus nN,lx_n-x_Kl < lx_n-x*llx_n -x_Kl + lx_K- x*l <.
Hence, X converges to x*.

Theorem : (Cauchy’s Principle of Convergence)

A necessary and sufficient condition for a series ∑u_n to converge is that to each ε > 0, there exists a positive integer m, such that |S_n - S_m| < ε
lu_m+1 + u_m+2+ ... + u_nI < ε, for all n ≥ m.
Proof: Let <S_n> be the sequence of partial sums of the series ∑u_n. By definition, ∑u_n converges ⇔ <S_n> converges
⇔ To each ε > 0, there exists a positive integer m such that
IS_n - S_ml < ε ∀ n ≥ m
(Cauchy's Principle of Convergence for Sequences)
⇔ I (u₁ + u₂ + ... + u_m + u_m+1, + u_m+2, + ... + u_n) - (u₁ + u₂ + ... + u_m)] < ε ∀ n ≥ m
⇔ |u_m+1, + u_m+2 + .... + u_n | < ε ∀ n ≥ m.
Hence the theorem.
Examples 1:

• The sequence  is a Cauchy sequence.
• The sequence <n²> is not a Cauchy sequence.
• The sequence <a_n> = <(-1)ⁿ> = < - 1, 1 , - 1 , 1, . . . ) is not a Cauchy sequence, for if <a_n> is a Cauchy sequence, then for ε = 1, there exists a positive integer m such that la_n - a_mI < 1 ∀ n ≥ m.
  If m is an even integer, then a_m = 1.
  We can choose n = 2m > m for which a_n = 1 and
  la_n - a_mI = |1 + 1| = 2 < 1, a contradiction.
  Hence ((-1)ⁿ> is not a Cauchy sequence.
• Consider the series (that is, infinite sum)
  
  We may view this as the limit of the sequence of partial sums
  
  We can show that the limit converges using Theorem by showing that {a_j} is a Cauchy sequence. Observe that if j, k > N, we definitely have
  
  It may be difficult to get an exact expression for the sum on the right, but it is easy to get an upper bound.
  
  The reason we used the slightly wasteful inequality, replacing 1/n² by 1/n²-n is that now the sum on the right telescopes, and we know it is exactly equal to 1/N-1. To sum up, we have shown that when j, k > N, we have
  
• Let 0 < α < 1 and let (x_n) satisfy the condition |x_n+1 - x_n| ≤ αⁿ or all n ∈ N then (x_n) is cauchy.
  Solution: For all m, n ∈ N with m > n, we have |x_m - x_n| ≤ |x_n - x_n+1| + |x_n+1 - x_n+2| + .... + |x_m-1 - x_m| ≤ αⁿ + αⁿ⁺¹ + ... + α^m-1 = αⁿ/1 - α (1 - α^m-n) < αⁿ/1-α. Since 0 < α < 1, α^m-n → 0 and so given any ε > 0, we can choose n_o ∈ N such that αⁿ/1-α (1 - 0) < ε. Hence for all m, n ≥ n_o, we have |x_m -x_n| < αⁿ/1-α (1 - 0) < ε. Therefore (x_n) is a Cauchy sequence.
• Let  or all n ∈ N then (x_n) is______
  Solution: For all m, n ∈ N with m > n, we have |x_m - x_n| =  Given ε > 0, we choose n_o ∈ N satisfying n_o > 2/ε. Then for all m, n ≥ n_o, we get |x_m - x_n| < 2/n_o < ε. Consequently (x_n) is a Cauchy sequence in R and hence (x_n) is convergent.
• Show that the sequence <a_n> defined by a_n = (—1 )ⁿ/n converges.
  Solution: If we show that <a_n> is a Cauchy sequence, then by Cauchy's General Principle of Convergence, <a_n> converges.
  Let ε > 0. If n ≥ m, then 1/n ≤ 1/m and
  
  
  Let m be a positive integer > 2/ε. Then
  |a_n - a_m| < ε ∀ n ≥ m.
  So <a_n> is a Cauchy sequence and hence convergent.

Example 2: (a) Show that if {a_n}^∞_n=1 is Cauchy then {a_n²}^∞_n=1 is also Cauchy.
(b) Given an example of a Cauchy sequence {a_n²}^∞_n=1 such that {a_n}^∞_n=1 is not Cauchy.
Solution: (a) Since {a_n}^∞_n=1 is Cauchy, it is convergent. Since the product of two convergent sequences is convergent the sequence {a_n²}^∞_n=1 is convergent and therefore is Cauchy.
(b) Let a_n = (-1)ⁿ for all n ∈ N. The sequence {a_n}^∞_n=1 is not Cauchy since it is divergent. However, the sequence {a_n²}^∞_n=1 = {1, 1, ....} converges to 1 so it is Cauchy.

Example 3: Let {a_n}^∞_n=1 be a Cauchy sequence such that an is an integer for all n ∈ N. Show that there is a positive integer N such that a_n = C for all n ≥ N, where C is a constant.
Solution: Let ε = 1/2. Since {a_n}^∞_n=1 is Cauchy, there is a positive integer N such that if m, n > N we have |a_m - a_n| < 1/2. But a_m - a_n is an integer so we must have a_n = a_N for all n ≥ N.

Example 4: What does it mean for a sequence {a_n}^∞_n=1 to not be Cauchy?
Solution: A sequence {a_n}^∞_n=1 is not a Cauchy sequence if there is a real number ε > 0 such that for all positive integers N there exist n, m ∈ N such that n, m ≥ N and la_n - a_m| < ε.

Example 5: Let {a_n}^∞_n=1 and {b_n}^∞_n=1 be two Cauchy sequences. Define c_n = la_n - b_nl. Show that {c_n}^∞_{n=1 is a Cauchy sequence.}
Solution: Let ε > 0 be given. There exist positive integers N₁ and N₂ such that if n, m ≥ N₁ and n, m ≥ N₂ we have la_n - a_mI < ε/2 and lb_n - b_ml < ε/2. Let N = N₁ + N₂. If n, m ≥ N then lc_n - c_ml = |la_n - b_n| — la_m - b_mI ≤ l(a_n - b_n) + (a_m - b_m)l ≤ la_n - a_mI + lb_n - b_mI < ε. Hence {c_n}^∞_n=1 is a Cauchy sequence.

Example 6: Explain why the sequence defined by a_n = (- 1)ⁿ is not a Cauchy sequence.
Solution: We know that every Cauchy sequence is convergent. We also know that the given sequence is divergent. Thus, it can not be Cauchy.

Example 7: Every subsequence of a Cauchy sequence is itself a Cauchy sequence.

For a sequence (s_n)_n∈N, a subsequence is a sequence of the form (s_nk)_k∈N for a strictly increasing sequence of natural numbers 1 ≤ n₁ < n₂ < n₃ < ..... For every integer k ∈ N, by induction on k, n_k is at least as large as k.
First, let (s_n)_n∈N, be a sequence that converges to s. Let (s_nk)_k∈N be a subsequence. For every ε > 0, since (s_n)_n∈N, converges, there exists N ∈ N such that for every n ∈ N with n ≥ N, |s_n - s| is less than ε. For every k ∈ N with k ≥ N, since n_k is at least as large as k, in particular n_k = N. Therefore, |s_nk -s| is less than ε. Thus (s_nk)_k∈N converges to s.
Next, let (s_nk)_k∈N be a Cauchy sequence. For every ε > 0, there exists N ∈ N such that for every m, n ∈ N with m ≥ N and with n ≥ N, |s_n - s_m| is less than ε. For every k, I ∈ N with k ≥ N and with I ≥ N, then n_k ≥ k ≥ N and n_i > I > N. Therefore, |S_nk - S_ni| is less than ε. Thus (s_nk)_k∈N  is a Cauchy sequence.

Example 8: If a subsequence of a Cauchy sequence converges to L, then the full sequence also converges to L.

Example 9: Prove directly from the definition that the sequence  is a Cauchy sequence.
Solution: Let ε > 0 be given. Let N be a positive integer to be chosen. Suppose that n, m ≥ N. We have



Example 10: Consider a sequence defined recursively by a₁ = 1 and a_n+1 = a_n + (-1 )ⁿ n³ for all n ∈ N. Show that such a sequence is not a Cauchy sequence. Does this sequence converge?
Solution: We will show that there is an ε > 0 such that for all N ∈ N there exist m and n such that m, n ≥ N but la_m - a_nl > ε. Note that la_n+1 - a_nI = n³ ≥ 1. Let N ∈ N. Choose m = N + 1 and n = N. In this case, la_m - a_nl = N³ ≥ 1 = ε. Hence, the given sequence is not a Cauchy sequence. Since every convergent sequence must be Cauchy, the given sequence is divergent.
Theorem : (Cauchy’s first theorem on limits)


Let ε > 0 be any number. Now  b = 0 implies that there exists a positive integer m such that
   ..... (iii)
Again  b_n = 0 ⇒ (b^) is convergent => <b_n> is bounded
⇒ there exists a number K > 0 such that |b_n| ≤ K ∀ n ∈ N.       .....(iv)

Let p be a positive integer > 2mK/ε. Then
  .... (v)
Now

 ... (vi)
Let q = max (m, p). Then   ... (vii)

Hence

Remark
The converse of the above theorem may not be true.
For example, <a_n> = <-1, 1, - 1, 1, ...) is not convergent
and a₁ + a₂ = 0, a₁ + a₂ + a₃ = - 1, a₁ + a₂ + a₃ + a₄ = 0, etc.
Now

Hence

Theorem : (Cauchy’s second theorem on limits)
If  a_n = I, where a_n > 0 ∀ n ∈ N, then  (a₁, a₂, ... a_n)^1/n = I.
Proof:

By Cauchy’s first theorem on limits,  (log a_n) = log I

Hence  (a₁, a₂ ... a_n)^1/n = I.
Corollary : Suppose <a_n> is a sequence of positive real numbers such that

Proof: 

By Theorem it follows that


Example 11: Prove that
Solution: We have


Using the Corollary of Theorem we get


Example 12: Prove that if  then <a_n> converges to 4/e.
Solution: Let

We have

Some important theorems on limits

Theorem : If an ≥ 0 ∀ n ∈ N and  an = I, then I ≥ 0.
Proof: Let, if possible, I < 0.
Let ε = —I > 0.
Since  a_n = I, so there exists a positive integer m such that
|a_n — I| < — I   ∀ n ≥ m
⇒ a_n - I ≤ |a_n - I| < — I ⇒ a_n < — I + I = 0  ∀ n ≥ m
⇒ a_n < 0    ∀ n ≥ m,
which contradicts the fact that a_n ≥ 0 ∀ n. Hence I ≥ 0
Theorem: If <a_n> and <b_n> be two sequences such that a_n ≤ b_n ∀ n ∈ N, then lim a_n ≤ lim b_n 
Proof: We see that a_n ≤ b_n ⇒ b_n ≥ a_n  ⇒ b_n - a_n ≥ 0
⇒ (b_n - a_n) ≥ 0,
⇒ b_n -  a_n ≥ 0 ⇒  b_n ≥ a_n.
Hence  a_n ≤  b_n.
Theorem : (Sandwich Theorem)

If <a_n>, <b_n> and <c_n> be three sequences such that
(i) a_n ≤ b_n ≤ c_n  ∀  n ∈ N,
(ii)  a_n = I =  c_n
then b_n = I.
Proof: Let ε > 0 be given.
Since b_n = I, there exists a positive integer m₁ such that

Since  c_n = I, there exists a positive integer m. such that

Let m = max {m₁, m₂}. Then
I - ε < a_n < I + ε   and  I - ε < c_n < I + ε, ∀ n ≥ m.      (*)
From (*) and condition (i), we obtain

Hence bn =I.
Theorem : Ratio test for convergence of sequence
If <a_n> be a sequence of positive number such that

• If I < 1, then (a_n) = 0
• If I < 1, then (a_n) = ∞
• If I = 1, then Test fails.

Proof: (1) Since I I I < 1, we can choose a positive number ε so small such that
I I I + ε + 1 or k < 1. where k = I I I + ε.
Since  there exists a positive integer m such that


Putting n = m, m + 1, ... , (n - 1) successively and multiplying, we get

We know kⁿ → 0 as n → ∞, since 0 < k < 1.
Using in (2), we see that Ia_nI → 0 as n → ∞. Hence lim a_n = 0.
Proof: (2) Since I > 1, we can choose a positive number ε such that
I - ε > 1 or k > 1, where k = I - ε.
Since  there exists a positive integer m such that

Consider


Putting n = m, m + 1, ... , (n - 1) successively and multiplying, we get

Now k > 1 ⇒ kⁿ → ∞ as n → ∞,
Hence
 a = ∞.
Proof: (3). Let a_n = 1 ∀ n
< 1 , 1, 1, . . . > → 1 (constant sequence)
(this is a convergent sequence)

Consider a_n = n ∀ n
< 1, 2, 3, 4, ... n, n + 1, . . . > → ∞
(this is a divergent sequence)

So in this case we cannot conclude anything about the sequence.

Example 1: Deduce that 2^-n n² = 0.
Solution:



Example 2: Show that (3^-nn³) = 0.
Solution: 



Example 3: Show that  for all y.
Solution: