Prove that $$\lim\limits_{x\to 0^+}\sum\limits_{n=1}^\infty\frac{(-1)^n}{n^x}=-\frac12$$

Does the limit even exist? Because when I develop the sum and consider the limit $x\to 0^+$, I get the sum $1-1+1-1+1-\dots$

  • $\begingroup$ Maple says $${\it polylog} \left( x,-1 \right) $$ $\endgroup$ – Dr. Sonnhard Graubner Oct 12 '17 at 9:45
  • $\begingroup$ answer is 1/2, but why? $\endgroup$ – Bakhytbek Oct 12 '17 at 9:53
  • 1
    $\begingroup$ $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{x}}$ really means $\lim_{m \to \infty}\sum_{n=1}^{m}\frac{(-1)^{n}}{n^{x}}$, so what you have is actually a double limit $\lim_{x\to 0^+}\lim_{m\to \infty}$. You can't simply swap the order of double limits to get $\lim_{m\to \infty}\lim_{x\to 0^+}$ (which is what gives you $-1+1-1+\cdots$) and expect to get the correct result. $\endgroup$ – Arthur Oct 12 '17 at 9:55
  • $\begingroup$ I'm sorry, yes the answer is -1/2 $\endgroup$ – Bakhytbek Oct 12 '17 at 9:58
  • $\begingroup$ After robjohn's answer, close to $x=0$, the expansion is $-\frac{1}{2}+\frac{1}{2}\log \left(\frac{2}{\pi }\right)x$ $\endgroup$ – Claude Leibovici Oct 12 '17 at 10:10

We can write this sum in terms of the Dirichlet eta function or the Riemann zeta function: $$ \begin{align} \sum_{n=1}^\infty\frac{(-1)^n}{n^z} &=-\eta(z)\\ &=-\left(1-2^{1-z}\right)\zeta(z)\tag1 \end{align} $$ In this answer it is shown that $\zeta(0)=-\frac12$. Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{z\to0}\sum_{n=1}^\infty\frac{(-1)^n}{n^z}=-\frac12}\tag2 $$

Computing $\boldsymbol{\eta'(0)}$

Using the formula $\eta(z)\Gamma(z)=\int_0^\infty\frac{\,t^{z-1}}{e^t+1}\,\mathrm{d}t$, we get $$ \begin{align} \eta(z)\Gamma(z+1) &=\int_0^\infty\frac{z\,t^{z-1}}{e^t+1}\,\mathrm{d}t\\ &=\int_0^\infty\frac1{e^t+1}\,\mathrm{d}t^z\\ &=\int_0^\infty\frac{t^z\,e^t}{\left(e^t+1\right)^2}\,\mathrm{d}t\tag3 \end{align} $$ As shown in this answer, $$ \int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=-\gamma\tag4 $$ Using Gautschi's Inequality, $\lim\limits_{n\to\infty}\frac{\Gamma(n+1)}{\Gamma\left(n+\frac12\right)} \frac{\Gamma(n+1)}{\Gamma\left(n+\frac32\right)}=1$; therefore, $$ \begin{align} \prod\limits_{k=1}^\infty\frac{2k}{2k-1}\frac{2k}{2k+1} &=\lim_{n\to\infty}\prod\limits_{k=1}^n\frac{k}{k-\frac12}\frac{k}{k+\frac12}\\ &=\lim_{n\to\infty}\frac{\Gamma(n+1)\,\color{#090}{\Gamma\left(\frac12\right)}}{\Gamma\left(n+\frac12\right)} \frac{\Gamma(n+1)\,\color{#090}{\Gamma\left(\frac32\right)}}{\Gamma\left(n+\frac32\right)}\\[6pt] &=\color{#090}{\frac\pi2}\tag5 \end{align} $$ Taking the derivative of $(3)$ and evaluating at $z=0$, we have $$ \begin{align} \eta'(0)-\frac12\gamma &=\int_0^\infty\log(t)\frac{e^{-t}}{(1+e^{-t})^2}\,\mathrm{d}t\tag{6a}\\ \eta'(0) &=-\int_0^\infty\log(t)\,\mathrm{d}\left(\frac{e^{-t}}{1+e^{-t}}-\frac{e^{-t}}2\right)\tag{6b}\\ &=-\frac12\int_0^\infty\log(t)\,\mathrm{d}\frac{e^{-t}(1-e^{-t})}{1+e^{-t}}\tag{6c}\\ &=\frac12\int_0^\infty\frac{e^{-t}(1-e^{-t})}{t\left(1+e^{-t}\right)}\,\mathrm{d}t\tag{6d}\\ &=\frac12\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty\frac{e^{-kt}-e^{-(k+1)t}}t\,\mathrm{d}t\tag{6e}\\ &=\frac12\sum_{k=1}^\infty(-1)^{k-1}\log\left(\frac{k+1}k\right)\tag{6f}\\ &=\frac12\sum_{k=1}^\infty\log\left(\frac{2k}{2k-1}\frac{2k}{2k+1}\right)\tag{6g}\\ &=\frac12\log\left(\frac\pi2\right)\tag{6h} \end{align} $$ Explanation:
$\text{(6a)}$: taking the derivative of $(3)$ and evaluating at $z=0$
$\text{(6b)}$: prepare to integrate by parts and subtract half of $(4)$ from both sides
$\text{(6c)}$: combine fractions
$\text{(6d)}$: integrate by parts
$\text{(6e)}$: write $\frac{e^{-t}}{1+e^{-t}}$ as a series in $e^{-t}$
$\text{(6f)}$: apply Frullani's Integral
$\text{(6g)}$: combine adjacent positive and negative terms
$\text{(6h)}$: apply $(5)$

Equation $(6)$ justifies Claude Leibovici's comment on the question that for $z$ slightly above $0$, $$ \sum_{n=1}^\infty\frac{(-1)^n}{n^z}\sim-\frac12-\frac z2\,\log\left(\frac\pi2\right)\tag7 $$

  • $\begingroup$ Thanks for posting material of so high quality ! Cheers. $\endgroup$ – Claude Leibovici Oct 13 '17 at 18:12
  • $\begingroup$ Another great answer in your characteristic style. +1 $\endgroup$ – Paramanand Singh Oct 14 '17 at 9:49

I'll look at $\sum_{n=1}^{\infty}(-1)^{n+1}/n^x$ instead. (I like starting with a positive summand!) We'll then want to show the limit is $1/2.$

This sum equals

$$\tag 1 \sum_{n=1}^{\infty}\left (\frac{1}{(2n-1)^x} - \frac{1}{(2n)^x} \right ) = \sum_{n=1}^{\infty}\frac{(2n)^x -(2n-1)^x }{(2n-1)^x(2n)^x}.$$

Now $(2n)^x -(2n-1)^x = xc^{x-1}$ by the MVT. Here $2n-1<c<2n.$ For $0<x<1$ we'll have $c^{x-1}> (2n)^{x-1}.$ It follows that $(1)$ is bounded below by

$$\tag 2\sum_{n=1}^{\infty}\frac{x}{(2n)^{1-x}(2n)^{x}(2n)^{x}} = x\sum_{n=1}^{\infty}\frac{1}{(2n)^{1+x}} > x \int_1^\infty (2t)^{-1-x} \, dt = x\cdot\frac{1}{x2^{1+x}}.$$

We get a bound from above using $c^{x-1}< (2n-1)^{x-1}.$ This shows $(1)$ is less than

$$\tag 3 x\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{1+x}} < x(1 + \int_1^\infty (2t-1)^{-1-x} \, dt) = x\left(1+\frac{1}{2x}\right).$$

As $x\to 0^+,$ the limits of the terms on the right of $(2)$ and $(3)$ are $1/2,$ and we're done.

  • $\begingroup$ This one use very simple tools compared to other answer. +1 and yes very few people would like a sum to begin with negative summand :) $\endgroup$ – Paramanand Singh Oct 14 '17 at 9:51
  • $\begingroup$ @ParamanandSingh Thanks. This approach seems far simpler to me, but it's elicited nothing but the sound of crickets so far. $\endgroup$ – zhw. Oct 15 '17 at 17:06
  • $\begingroup$ Not to be disappointed. Voting is based on many things and perhaps everyone may not value simplicity in the way I do. $\endgroup$ – Paramanand Singh Oct 15 '17 at 17:23

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{x \to 0^{\large +}}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^x} = -\,{1 \over 2}:\ {\large ?}}$

\begin{align} \lim_{x \to 0^{\large +}}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^x} & = \lim_{x \to 0^{\large +}}\mrm{Li}_{x}\pars{-1} \end{align}

$\ds{\mrm{Li}_{s}}$ is the Order-$\ds{s}$ Polylogarithm.

By using its integral representation: \begin{align} \lim_{x \to 0^{\large +}}\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^x} & = -\lim_{x \to 0^{\large +}}\bracks{% {1 \over \Gamma\pars{x}}\int_{0}^{\infty}{t^{x - 1} \over \expo{t} + 1}\,\dd t} \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, -\lim_{x \to 0^{\large +}}\braces{% {1 \over \Gamma\pars{x}}\bracks{% {1 \over x}\int_{0}^{\infty}{t^{x}\expo{t} \over \pars{\expo{t} + 1}^{2}} \,\dd t}} \\[5mm] & = -\lim_{x \to 0^{\large +}}\bracks{% {1 \over \Gamma\pars{x + 1}} \int_{0}^{\infty}{t^{x}\expo{t} \over \pars{\expo{t} + 1}^{2}}\,\dd t} = -\int_{0}^{\infty}{\expo{t} \over \pars{\expo{t} + 1}^{2}}\,\dd t \\[5mm] & = \left.{1 \over \expo{t} + 1}\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} = \bbx{-\,{1 \over 2}} \end{align}


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