# How do I test for convergence of $\Sigma_{n = 2}^{\infty} \frac{log(n)}{n \sqrt{n + 1}}$

I was trying to solve this problem.

Test $$\Sigma_{n = 2}^{\infty} \frac{log(n)}{n \sqrt{n + 1}}$$ for convergence or divergence .

But I couldn't quite make a lot of progress. Here's what I tried.

• I tried applying integral test $$\int_{2}^{\infty} \frac{log(x)}{x \sqrt{x + 1}}dx$$ and proving that this series is bounded but I couldn't solve the integral.
• The next thing that I thought of was applying the direct comparison test where $$\frac{log(n)}{n \sqrt{n + 1}} < \frac{log(n)}{n}$$. I tried figuring out the convergence of $$\Sigma_{n = 2}^{\infty}\frac{log(n)}{n}$$ but this series diverges.
• Next I again tried applying direct comparison test with $$\frac{log(n)}{n \sqrt{n + 1}} < \frac{log(n)}{\sqrt{n+ 1}}$$ and tried applying integral test for $$\Sigma_{n = 2}^{\infty} \frac{log(n)}{\sqrt{n+ 1}}$$ but while solving for this integral using integration by-parts technique I could see that this series is also divergent.

Can anyone help me out with this problem? Preferably using only direct comparison test, limit comparison test, and integral test.

Thanks!

$$\sum_{n = 2}^{\infty} \frac{log(n)}{n \sqrt{n + 1}}$$
The basic fact needed is that $$\dfrac{\ln(n)}{n^a} \to 0$$ as $$n \to \infty$$ for any $$a > 0$$.
Setting $$a= 1/4$$, $$\dfrac{\ln(n)}{n^{1/4}} \to 0$$ so that $$\dfrac{\ln(n)}{n^{1/4}} \lt 1$$ for all large enough $$n$$.
Therefore, for all large enough $$n$$, $$\dfrac{\ln(n)}{n\sqrt{n+1}} \lt \dfrac{n^{1/4}}{n\sqrt{n+1}} \lt \dfrac1{n^{5/4}}$$ and the sum of these converges.
• Did you think of setting $n = \frac{1}{4}$ with the intention of converting the entire series into a p-series ($\Sigma \frac{1}{n^p}$) with $p > 1$ ? May 11 '19 at 5:19