Determine whether $\sum_{1}^{\infty} (\frac{\sqrt{2n-1} \cdot \log(4n+1)}{n(n+1)})$ is converge.

Question

I have a question about infinite series;

Determine whether the series are converges or diverges

$\sum_{1}^{\infty} (\frac{\sqrt{2n-1}\cdot \log(4n+1)}{n(n+1)})$

I've just seen integral and comparison test so far, so if the answers just contain these methods, it would be nice.

My attempts;

I neither be able to integrate the function nor find a proper function to compare, but I thought if I prove that when n goes $\infty$, the function goes to 0 and the function's first derivative is less than 0 for all n in $(0,\infty)$, then the series converge.

So what do you guys think about my thought about solving the program ?, and what is the proper solution of this problem ?

Answer 1

This series looks a lot like: $\sum_1^\infty \frac{\sqrt{n} \cdot \log(n)}{n^2}$, so we will compare against this series.

By the limit comparison test,

$$\lim_{n \to \infty} \frac{\frac{\sqrt{2n-1} \cdot \log(4n+1)}{n(n+1)}}{\frac{\sqrt{n} \cdot \log(n)}{n^2}} = \lim_{n \to \infty} \frac{\sqrt{2n-1} \cdot \log(4n+1)}{n(n+1)} \cdot \frac{n^2}{\sqrt{n} \cdot \log(n)}$$$$ = \lim_{n \to \infty} \frac{\sqrt{2n-1}}{\sqrt{n}}\cdot \frac{n^2}{n(n+1)} \cdot \frac{\log(4n+1)}{\log(n)} = \sqrt{2} \cdot 1 \cdot 1 = \sqrt{2},$$

which is positive and finite, thus the original series converges if and only if $\sum_1^\infty \frac{\sqrt{n} \cdot \log(n)}{n^2}$ converges. To show this series converges, you can use the direct comparison test and the fact that $\log(n) \leq \sqrt{n}$ for large enough $n$.

Answer 2

hint: $\log n \leq \sqrt[4]{n}$, for $n \ge N_0$.