Convergence of $\sum_{n=1}^{\infty}\left(\frac{n}{n^2+1}\right)^{k(n)}\,\,\,\,\,;\,\,k(n)=\frac{1}{\cos\left(\frac{1}{\ln^{a}(n)}\right)}$

Question

Study the convergence of the series as $a > 0$

$$\sum_{n=1}^{\infty}\left(\frac{n}{n^2+1}\right)^{k(n)}\,\,\,\,\,;\,\,k(n)=\frac{1}{\cos\left(\frac{1}{\ln^{a}(n)}\right)}$$

In the text there's the integer part of $\left(\frac{n}{n^2+1}\right)$, but I think it is a typo beacuse it would be identically zero. How to handle $k(n)$ in this case?

Answer 1

We have that

$$\cos\left(\frac{1}{\ln^{a}(n)}\right)\sim 1-\frac{1}{2\ln^{2a}(n)}$$

$$k(n)=\frac{1}{\cos\left(\frac{1}{\ln^{a}(n)}\right)}\sim 1+\frac{1}{2\ln^{2a}(n)}$$

therefore

$$\left(\frac{n}{n^2+1}\right)^{k(n)} \sim \frac1n$$

Answer 2

Hint First you can write :

$$\left(\frac{n}{n^2+1}\right)^{k(n)} = \exp \left(k(n)\log\left(\frac{n}{n^2+1}\right)\right) $$

And you can try to approximate $k(n)$ and $\log\left(\frac{n}{n^2+1}\right)$ as $n \rightarrow \infty$.