Is the given series convergent?

Question

The infinite series $\sum\limits_{n=1}^\infty \frac {a^{n} \log n} {n^{2} } $ converges if and only if

(A) $a ∈ [−1, 1)$

(B) $a ∈ (−1, 1]$

(C) $a ∈ [−1, 1]$

(D) $a ∈ (−∞, ∞)$

My attempt: The correct answer is option A) because if $a=1$, $$\lim_{n\to ∞}a^{n}= ∞ .$$ So the correct option is A) that is $a ∈ [−1, 1)$.

Is my answer is correct or not, please verify and tell me the solution, I would be more than thankful.

@ Henry,,im not getting ur idea,,,how it will converges to 1 — jasmine, Oct 30 '17 at 08:19
If $a=1$ then $a^n=1$ for all $n$ and you are looking at $\sum_{n=1}^\infty \frac { \log n} {n^{2} }$ which does converge, in fact to about $0.937548$. So option A is wrong. — Henry, Oct 30 '17 at 12:28

Chinnapparaj R · Accepted Answer · 2017-10-30T09:04:14.283

1

Let us use the ratio test :

$$ \left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{a^{n+1}\log(n+1)}{(n+1)^2}.\frac{n^2}{a^n \log n} \right|$$

$$ =\left| a.\frac{\log (n+1)}{\log n}.\Big(\frac{n}{n+1}\Big)^2 \right|$$

$$\rightarrow \vert a \vert$$ so the given series converges if $\vert a \vert<1.$

Also if $a=1$, then then given series becomes $\sum \frac{\log n}{n^2}$, which is converges(see here )

If $a=-1$, then then given series becomes $\sum \frac{(-1)^n\log n}{n^2}$, which is absolutely converges and hence converges

so the answer is $a\in [-1,1]$

edited Oct 30 '17 at 09:04

answered Oct 30 '17 at 08:43

Chinnapparaj R

thanks a lot @ chinnapparaj but liitle bit confuse $$\lim_{n\to ∞}a^{n}= ∞ $$ if a = 1...if a=1 and $1^{∞} = ∞$ – jasmine Oct 30 '17 at 08:57
1

when $a\in (-1,1)$, then $a^n \rightarrow 0$ as $n \rightarrow \infty$. When $a=1$, the $a^n=1^n=1$, a constant sequence which tends to $1$ – Chinnapparaj R Oct 30 '17 at 09:00
thanks a lot once again@ Chinnapparaj R – jasmine Oct 30 '17 at 09:41

1 Answers1