Evaluating the infinite and partial sum of an alternating series

Question

The inspiration comes from a physical problem. Let's focus on the non-negative half of the number axis -- so we are studying a 1D system. We put 1 unit positve charge on "site" 0, then a unit negative charge on site 1, then a unit positive charge on site 2, ... See figure below.

The question is what's the force on the zeroth charge (the left most one). More specifically,

1. what's the total force if we are considering all the charges. (the infinite sum)
2. what's the partial force if we are considering forces exerted by the first N charges, here N excludes the 0th charge. (the partial sum)

The physics is just Columb's law. We neglect all physical constants, and since all charges are unit, we have the force expressed as a series

$$ F(N) = \sum_{n=1}^N \frac{(-1)^{n-1}}{n^2} $$

A plot in Mathematica shows $F(100)$ is well converged and is about $0.82242$. But I am sure someone has more elegant analysis. I myself did a lousy approximation by "spreading" a charge into an interval as $$ \frac{\pi}{2} {\cos(\pi x)} $$ $F(N)$ is then approximated by an integration. $$ F(N) \sim -\frac{\pi}{2} \int_{1/2}^{N+1/2} \frac{\cos(\pi x)}{x^2} dx $$ This approximation is very crude and for the infinite sum, it gives $F(\infty) \sim 0.98713$ - not bad, but not good either.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{F}\pars{N} & \equiv \sum_{n = 1}^{N}{\pars{-1}^{n - 1} \over n^{2}} = \sum_{n = 1}^{\infty}{\pars{-1}^{n - 1} \over n^{2}} - \sum_{n = 1}^{\infty}{\pars{-1}^{n + N - 1} \over \pars{n + N}^{2}} \\[5mm] & = -\,\mrm{Li}\pars{-1} - \pars{-1}^{N} \braces{{1 \over 4}\bracks{\zeta\pars{2,{1 \over 2}\,N + {1 \over 2}} - \zeta\pars{2,{1 \over 2}\,N + 1}}} \\[5mm] & = {\pi \over 12} - {1 \over 4}\,\pars{-1}^{N} \bracks{\zeta\pars{2,{1 \over 2}\,N + {1 \over 2}} - \zeta\pars{2,{1 \over 2}\,N + 1}} \end{align}

$\ds{\zeta\pars{s,a}}$ is the Hurwitz Zeta Function. It has the following asymptotic behaviour:

$$ \zeta\pars{2,a} \sim {1 \over a}\quad\mbox{as}\ a \to \infty $$

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Then, \begin{align} \mrm{F}\pars{N} & \equiv \sum_{n = 1}^{N}{\pars{-1}^{n - 1} \over n^{2}} \sim {\pi^{2} \over 12} - {1 \over 4}\,\pars{-1}^{N}\pars{{1 \over N/2 + 1/2} - {1 \over N/2 + 1}} \\[5mm] & \sim \bbx{\ds{{\pi^{2} \over 12} - {\pars{-1}^{N} \over 2N^{2}}\quad\mbox{as}\ N \to \infty}} \end{align}

Answer 2

This is too long for a comment.

Felix Marin gave the solution for the asymptotics. I think that we could have very good approximations of the partial sums using the expansion $$\zeta(2,a)=\frac{1}{a}+\frac{1}{2 a^2}+\frac{1}{6 a^3}-\frac{1}{30 a^5}+\frac{1}{42 a^7}+O\left(\frac{1}{a^8}\right)$$ which would lead to $$F(N) = \sum_{n=1}^N \frac{(-1)^{n-1}}{n^2}\approx \frac{\pi ^2}{12}-(-1)^N\frac{ \left(N^5-N^4+N^2-3\right)}{2 N^7}$$ Below is produced a short table for rather small values of $N$ $$\left( \begin{array}{ccc} N & \text{exact} & \text{approximation} \\ 2 & 0.7500000000 & 0.7560607834 \\ 3 & 0.8611111111 & 0.8608758126 \\ 4 & 0.7986111111 & 0.7986328049 \\ 5 & 0.8386111111 & 0.8386078334 \\ 6 & 0.8108333333 & 0.8108340173 \\ 7 & 0.8312414966 & 0.8312413172 \\ 8 & 0.8156164966 & 0.8156165524 \\ 9 & 0.8279621756 & 0.8279621558 \\ 10 & 0.8179621756 & 0.8179621834 \\ 11 & 0.8262266384 & 0.8262266351 \\ 12 & 0.8192821940 & 0.8192821955 \\ 13 & 0.8251993537 & 0.8251993530 \\ 14 & 0.8200973129 & 0.8200973133 \\ 15 & 0.8245417574 & 0.8245417572 \\ 16 & 0.8206355074 & 0.8206355075 \\ 17 & 0.8240957150 & 0.8240957149 \\ 18 & 0.8210092952 & 0.8210092953 \\ 19 & 0.8237793783 & 0.8237793783 \\ 20 & 0.8212793783 & 0.8212793783 \end{array} \right)$$

For $N=100$, the difference between exact and approximated values is $1.11\times 10^{-16}$.

Answer 3

Here is another way to evaluate the series $S=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}$ in closed form without appealing to special functions or complex-plane analysis. Rather, we transform the series to an integral that we can evaluate using standard methodologies.

We begin by recasting $S$ as

$$\begin{align} \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}&=\sum_{n=1}^\infty\left(\frac{1}{(2n-1)^2}-\frac1{(2n)^2}\right)\\\\ &=\sum_{n=1}^\infty\left(\frac{1}{(2n-1)^2}+\frac1{(2n)^2}\right)-\frac12\sum_{n=1}^\infty\frac{1}{n^2}\\\\ &=\frac12\sum_{n=1}^\infty\frac{1}{n^2}\tag 1 \end{align}$$

Next, we make use of the identity $\frac1n=\int_0^1x^{n-1}\,dx$ to write

$$\begin{align} \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}&=\frac12\sum_{n=1}^\infty\int_0^1x^{n-1}\,dx\int_0^1y^{n-1}\,dy\\\\ &=\frac12\int_0^1\int_0^1\sum_{n=1}^\infty(xy)^{n-1}\,dx\,dy\\\\ &=\frac12\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy\tag 2 \end{align}$$

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In THIS ANSWER, I used the straightforward coordinate transformation $x=s+t$, $y=s-t$ to evaluate the integral on the right-hand side of $(2)$.

This transformation leads to integrals for which anti-derivatives can be written in terms of elementary functions. The final result is

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}}$$

as was expected!