# 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:

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?

• That is what you need to show, yes. Best to be explicit about what $a_n$ is from the start. You obviously can't just assert (2) and (3), but have to prove them. – Thomas Andrews Jan 23 '13 at 14:28
• 1) and 2) are obvious; 3) it's proved with some passages and with passage to the limit. But wolfram said: "sum does not converge, ratio test inconclusive, root test inconclusive" – user55114 Jan 23 '13 at 14:33
• How can you prove (3) by passing to the limit, when you are trying to prove it for specific values, or do you just mean the function is strictly dereasing as a continuous function? – Thomas Andrews Jan 23 '13 at 14:36
• you can arrive to $log^2(n)> n*log^2(1+1/n) and with passage to limit the first part --> +inf, the second part of inequality -->0 so definitively I have proved the inequality – user55114 Jan 23 '13 at 14:58 • Again, you say "passage to the limit." Part of it is that you misformatted that comment, but I also just don't understand the thrust of your argument. What limit? If you send$n\to\infty$how are you showing that$\log(n)/\sqrt n >\log(n+1)/\sqrt{n+1}$? – Thomas Andrews Jan 23 '13 at 15:18 ## 3 Answers 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• Yes, then the series converges, right? – user55114 Jan 23 '13 at 14:50 • @Giacomo By the alternating series (Leibniz) test, it does – Nameless Jan 23 '13 at 14:52 • ok, thank you. But why wolfram said "does not converge" ? – user55114 Jan 23 '13 at 14:53 • @Agenog Wolfram Alpha says it converges. – user21467 Apr 5 '14 at 17:21 • @Agenog I also read the answer but I didn't see the conclusion. It happens frequently in MSE. Anyway, the answer is pretty fine. – Felix Marin Sep 27 '16 at 21:17 $$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ Indeed, it can be explicitly evaluated such that a 'closed expression' exists. Note that: 1. \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} 2. 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} 3. 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} Yes, your solution is correct. An alternative solution using the mean value theorem: Note that ifa_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.