On the convergence of a sequence

Question

Question: The infinite series $$ \sum_{n=1}^{\infty} \frac{a^n \log_e n}{n^2}$$ converges if and only if

A. $a \in [-1,1)$

B. $a \in (-1,1]$

C. $a \in [-1,1]$

D. $a \in (-\infty, \infty)$

My Approach:

We use the Ratio Test to test for convergence.

$$\lim_{n\to\infty} |t_n|^{1/n} = \lim_{n\to\infty} |\frac{a^n \log_e n}{n^2}|^{1/n}= \lim_{n\to\infty} |\frac{a (\log_e n^{1/n})^{1/n}}{n^{1/n}}|$$

Now from the knowledge that $\lim_{n\to\infty} n^{1/n}=1$, we have

$$ \lim_{n\to\infty} |t_n|^{1/n} = 0 $$

since $\lim_{n\to\infty} (\log_e n^{1/n})^{1/n} = 0 $

We see that $\lim_{n\to\infty} |t_n|^{1/n} < 1 $. Hence, the series is always convergent and its convergence does not depend on the value of $a$, thus $a$ can take any value and Option (D) is correct.

Is this line of reasoning correct? What other (preferably shorter) method could I have proceeded to arrive at this answer?

Answer 1

Note that by ratio test

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

and by root test

$$\left|\frac{a^n \log n}{n^2}\right|^{1/n}=|a|\left|\frac{(\log n)^{1/n}}{(n^2)^{1/n}}\right|\to |a|$$

indeed

• $(\log n)^{1/n}=e^{\frac{\log\log n}{n}}\to e^0=1$

• $(n^2)^{1/n}=e^{\frac{2\log n}{n}}\to e^0=1$

For the boundaries cases

• $a=-1$ refer to alternate series test

• $a=1$ refer to limit comparison test with $\sum \frac1{n^{3/2}}$

Answer 2

Basically what you said is $\lim\limits_{n \rightarrow \infty} (\log_e(n^{1/n}))^{1/n}= \lim\limits_{n \rightarrow \infty} \Big(\lim\limits_{n \rightarrow \infty} \log_e(n^{1/n})\Big)^{1/n} = 0$ but that's wrong.

For instance let's take rhe sequence $\Big((1+\frac{1}{n})^{1/n}\Big)_{n \in \mathbb{N}}$. You might know that this sequence converges to $e$ and not $1$.

Here you'd have to use the exponential form to find the limit: $(\ln(n))^{1/n} = e^{\frac{1}{n}\ln(ln(n))} \le e^{\frac{1}{n}\ln(1+n)} \rightarrow e^0 = 1$

Then $\lim\limits_{n \rightarrow \infty} \mid t_n^{1/n} \mid = \mid a \mid$

Other ways to test for convergence is to use ratio test or a comparison with a Riemmann series. For instance $n^{0.5}t_n = a^n\ln(n)$ which converges to 0 if and only if $-1 \le a \le 1$.