Sequences And Series Of Real Numbers -1 - IIT JAM MCQ

Here lower bound = l = − 1/2 , & upper bounded u = 1. But the given sequence is neither increasing nor decreasing. Thus, <s_n> is bounded but not monotonic.

The sequence of n th partial sum of given series is unbounded and goes to infinity.
since sequence of n th partial sum is unbounded, it is not convergent therefore the given series
is not convergent i.e. divergent.

If ∑μ_n be the given series, then we have


∴ From ratio test, the given series ∑μ_n is convergent or divergent according as 

Hence, the answer is option C

The ratio of consecutive numbers is close to 3. Comparing these terms with the sequence of {3ⁿ} which is 3, 9, 27 …. Comparing these terms with the corresponding terms of sequence {3ⁿ} and the nth term is 2 less than the corresponding power of 3.

R = 1 / P = 1 = radius of convergence
When x = 1 & x = –1, Then is converges at both 

x = 1 & x = – 1.
so, its interval of convergence is exactly [–1, 1].

We have 

is also convergent series.
∴ u_n is also convergent series.

Hence, by d’Alembert test 

implies u_n is divergent series.

By Cauchy condensation test, ∑μ_n is convergent series.
(D) 
By Leibnitz’s test, the series is convergent.
Also,  is divergent series. 
So, the given series is converges conditionally.

Here it is given that

Now, taking auxiliary series 

Which is finite and non-zero. Since, v_n is divergent.
Hence, t_n is also divergent.

Here given series is  is convergent

∴ The given series will be convergent, if



Hence, v_n is convergent series, so u_n is also convergent.
∴ The given series is convergent, if | x | ≤ 1.

In the above we have used the result that 1/n → 0 as n → ∞ and a result about combining convergent sequences and noting that the denominator converges to a non-zero value.

which is finite and non-zero.
Since, ∑v_n is divergent, therefore, ∑u_n is also divergent.

Hence, by Cauchy’s root test, the series convergent.

Here, it is given that 

So <S_n> is monotonically increasing Next, we will show that it is bounded.

⇒ <S_n> is bounded. By the theorem's Every monotonic bounded sequence is convergent, Then <S_n> is convergent.
By the Lemma, if <S_n> is a convergent sequence of real numbers, Then <S_n> is a Cauchy-sequence.
 

For limit n tending to infinity the sum also tends to infinity and thus it is not summable.

The nth term of the given series is

Hence, ∑u_n is convergent.

The series is conditionally convergent.