[Math] Does the series $\sum\limits_{n=1}^\infty\frac{\ (-1)^n \log(n)}{\sqrt{n}}$ converge

convergence-divergencesequences-and-series

The following series (OEIS A265162) converge or diverge?

$$\sum_{n=1}^\infty\frac{\ (-1)^n \log(n)}{\sqrt{n}}$$

I have proved that this series diverges absolutely.

I tried to use Leibniz criterion:

  1. $a_n >0$ definitively.
  2. The limit of $a_n=0$ (as n tends to infinity).
  3. $\log(n)/\sqrt n >\log(n+1)/\sqrt{n+1}$ definitively

it's ok?

Best Answer

As you said 1. is obvious.

For 2., by De L'Hospital, $$\lim_{n\to +\infty}\frac{\log n}{\sqrt n}=\lim_{n\to +\infty}\frac{\frac 1n}{\frac1{2\sqrt n}}=\lim_{n\to +\infty}\frac{2\sqrt n}n=0$$

For 3. you can either proceed with induction or show $f(x)=\frac{\log x}{\sqrt{x}}$ is stricly decreasing in $(N,+\infty)$ (choose $N$ sufficiently large). Indeed, $$f'(x)=\frac{\frac1 x\sqrt{x}-\frac{\log x}{2\sqrt{x}}}{x}$$ and $$\frac1 x\sqrt{x}-\frac{\log x}{2\sqrt{x}}=\frac{2-\log x}{\sqrt{x}}<0$$ for $x>e^2$

Related Question