# 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 $p > 1$ ? – Deepam Sarmah May 11 '19 at 5:19
• 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. – marty cohen May 11 '19 at 6:49

$$\frac{\log{n}}{n\sqrt{n+1}} \leq \frac{1}{n\sqrt{n}}$$

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

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

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?