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