I was trying to solve this problem.
Test $\sum\limits_{n = 2}^{\infty} \frac{\log(n)}{n \sqrt{n + 1}}$ forconvergence 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 $\sum\limits_{n = 2}^{\infty}\frac{\log(n)}{n}$ 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 $\sum\limits_{n = 2}^{\infty} \frac{\log(n)}{\sqrt{n+ 1}}$ 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!