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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!

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$\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.

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  • $\begingroup$ 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 $p > 1$ ? $\endgroup$ May 11 '19 at 5:19
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    $\begingroup$ The denominator has p=3/2 so I needed some a so that p-a > 1. a=1/4 was the simplest. Also, I see a minor error to fix. $\endgroup$ May 11 '19 at 6:49

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