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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 $p > 1$ ? $\endgroup$ – Deepam Sarmah 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. $\endgroup$ – marty cohen May 11 '19 at 6:49
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$\frac{\log{n}}{n\sqrt{n+1}} \leq \frac{1}{n\sqrt{n}} $

And it is easy to prove that the integral $\int_{1}^{\infty}\frac{1}{x\sqrt{x}} dx$ converges

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  • $\begingroup$ $\frac{\log{n}}{n\sqrt{n+1}} \leq \frac{1}{n\sqrt{n}}$ does not hold for all $n\in\mathbb N$. $\endgroup$ – Hirshy May 10 '19 at 20:02
  • $\begingroup$ Yeah, that's incorrect. I just plugged it into desmos. $\endgroup$ – Deepam Sarmah May 10 '19 at 20:04
  • $\begingroup$ No, it is not, it is only true for $0<n<3.15112...$. This can easily be seen: $$ \frac{\log{n}}{n\sqrt{n+1}} \leq \frac{1}{n\sqrt{n}} \Leftrightarrow \sqrt{n}\cdot \log(n)\leq \sqrt{n+1} \Leftrightarrow \ln(n) \leq \sqrt{1+\frac{1}{n}}$$ $\endgroup$ – Hirshy May 10 '19 at 20:12
  • $\begingroup$ Woops, mental fart. Use the fact that $\log{n} \leq Cx^{\alpha} $, for n big enough and $alpha > 0$ $\endgroup$ – user3646987 May 10 '19 at 20:12
  • $\begingroup$ Then you get that $\frac{\log n}{n \sqrt{n+1} }\leq C\frac{n^{\alpha} } { n \sqrt{n+1} }$ And choose $alpha < 0.5$ $\endgroup$ – user3646987 May 10 '19 at 20:14
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Hint:

We see that $$ \frac{\log(n)}{n\sqrt{n+1}} = \frac{\ln(n)}{\ln(10)n\sqrt{n+1}} \sim \frac{1}{\ln(10)n^{3/2}\ln(n)^{-1}} $$ which has the same behaviour as the improper integral of $$ f : x \longmapsto \frac{1}{x^{3/2}\ln(x)^{-1}} $$ Can you finish?

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