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


*

*$a_n >0$  definitively.

*The limit of $a_n=0$ (as n tends to infinity).

*$\log(n)/\sqrt n >\log(n+1)/\sqrt{n+1}$ definitively 


it's ok?
 A: 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 $x>e^2$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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Indeed, it can be explicitly evaluated such that a 'closed expression' exists.

Note that:


*

*\begin{align}
\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{s}} & =
\pars{2^{s} - 2}\zeta\pars{s} - 2^{-s}\qquad
\pars{~\zeta:\ \text{Zeta Function}~}
\end{align}

*Derive both members respect of $\ds{s}$:
\begin{align}
&-\sum_{n = 1}^{\infty}{\pars{-1}^{n}\ln\pars{n} \over n^{s}}
\\[5mm] = &\
-\ln\pars{2}\zeta\pars{s} + \ln\pars{2}\pars{1 - 2^{1 - s}}\zeta\pars{s} -
\pars{1 - 2^{1 - s}}\zeta'\pars{s}
\end{align}

*Take the limit $\ds{s \to 1/2}$:
\begin{align}
&\color{#f00}{\sum_{n = 1}^{\infty}{\pars{-1}^{n}\ln\pars{n} \over \root{n}}} =
\root{2}\ln\pars{2}\zeta\pars{1 \over 2} -
\pars{\root{2} - 1}\zeta'\pars{1 \over 2}
\\[5mm] = &\
\color{#f00}{\braces{\root{2}\ln\pars{2} - {1 \over 4}\pars{\root{2} - 1}
\bracks{\vphantom{\Large A}2\gamma + \pi +
2\ln\pars{8\pi}}}\zeta\pars{1 \over 2}}
\\[5mm] & \approx 0.1933\qquad
\pars{~\gamma:\ \text{Euler-Mascheroni Constant}~}
\end{align}
A: Yes, your solution is correct.
An alternative solution using the mean value theorem:
Note that if $a_n \to_{n\to \infty} 0$ then $\sum_{n \ge 0 } (-1)^n a_n$ and $\sum_{n \ge 0 }  (a_{2n} - a_{2n-1})$ either both converge or both diverge. Set $a_n=\frac{ \ln n}{n^{3/2}}$.
By the mean value theorem $a_{2n} - a_{2n-1}= \frac{2-\log(t)}{t^{3/2}}$ for some $t \in ]2n-1;2n[$.
Now notice that because $\frac{\log q}{q^{1/4}} \to 0$ there exists $q_0$ such that  $(\forall q>q_0),(\ln q<q^{1/4})$ .
Therefore $|\frac{2-\log(t)}{t^{3/2}}| \le \frac{2+\log(t)}{t^{3/2}} \le \frac{2}{t^{3/2}} +\frac{1}{t^{5/4}} $.
Having this, it is simple to determine the convergence of the original problem.
