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Could someone please help me with what is the sum of the $$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{4n^{2}-1}$$ infinite series? You can find the sum of $$\sum_{n=1}^{\infty}\frac{1}{4n^{2}-1}$$ here. Thank you.

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hint Per the other one, it suffices to compute $$ -\sum_{n=1}^\infty (-1)^n\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right),$$ (and then multiply by $-1/2$) but now we lose the nice telescoping property and instead have $$ \left(1 - \frac{1}{3}\right) -\left(\frac{1}{3} - \frac{1}{5}\right) +\left(\frac{1}{5} - \frac{1}{7}\right)-\ldots = 1 -\frac{2}{3}+\frac{2}{5} - \frac{2}{7}\ldots. $$ But the sum $$ 1-\frac{1}{3} +\frac{1}{5}-\frac{1}{7}+\ldots $$ is well known and closely related.

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We have $$ 1-\frac{2}{3} + \frac{2}{5} -\frac{2}{7} = -1 + 2\left(1-\frac{1}{3} + \frac{1}{5}-\ldots \right) = -1+\frac{\pi}{2}.$$ Because the series I opted to start with is your series times $-2,$ the final answer is $-1/2$ times this: $\pi/4-2.$

The formula $$ 1-\frac{1}{3} + \frac{1}{5}-\ldots = \frac{\pi}{4}$$ follows from the Taylor series $$ \arctan(x) = x-\frac{1}{3}x^3+\frac{1}{5}x^5 -\ldots$$ evaluated at $x=1.$ So my answer much different in substance from Olivier's.

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  • $\begingroup$ Wait... isn’t that equal to $\dfrac \pi 4$ ?? I could be wrong, because after $\dfrac 17$ there should be a $+$ operation instead of $-$ since it is a constant alternating sum. $\endgroup$ – Mr Pie Jan 17 '18 at 1:57
  • $\begingroup$ @user477343 The last sum I wrote down is $\pi/4,$ from which it follows that OP's sum is $1/2-\pi/4.$ (And you're right that I had a typo with the last sign.) $\endgroup$ – spaceisdarkgreen Jan 17 '18 at 2:06
  • $\begingroup$ Could you please extend your answer with more details? It would be very nice of you. By the way why is it well known that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n+1}=\frac{\pi}{4}$? $\endgroup$ – bencemeszaros Jan 17 '18 at 2:26
  • $\begingroup$ @bencemeszaros Well that's Leibniz's formula, isn't it? Discovered in $1673$ CE I believe... although it was discovered earlier $\approx 1400$ CE..... Oh wait, I though you were asking me XD $\endgroup$ – Mr Pie Jan 17 '18 at 2:50
  • $\begingroup$ @bencemeszaros k updated. User gave the history lesson. I believe it was first introduced and motivated to me as a horrendously slow and inefficient way to compute $\pi.$ $\endgroup$ – spaceisdarkgreen Jan 17 '18 at 3:22
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Assume $|x|<1$. One may recall that $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{2n-1}}{2n-1}=\arctan x $$then writing $$ \begin{align} 2\sum_{n=1}^\infty\frac{(-1)^{n}x^{2n-1}}{4n^{2}-1}&=\sum_{n=1}^\infty\frac{(-1)^{n}x^{2n-1}}{2n-1}-\sum_{n=1}^\infty\frac{(-1)^{n}x^{2n-1}}{2n+1} \end{align} $$ gives

$$ \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n}x^{2n-1}}{4n^{2}-1}&=\frac1{2x}-\frac12\left(\frac1{x^2}+1\right)\cdot\arctan x,\qquad 0<|x|<1. \end{align} $$

In particular, by letting $x \to 1^-$, one finds

$$ \sum_{n=1}^\infty\frac{(-1)^{n}}{4n^{2}-1}=\frac12-\frac \pi4 $$

using $\displaystyle \arctan 1= \frac \pi4$.

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  • $\begingroup$ Thank you for the answer. I accepted the other one because it was more intuitive for me. $\endgroup$ – bencemeszaros Jan 17 '18 at 10:56
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Or using our favourite alternative of converting the sum into a double integral.

As $$\sum_{n = 1}^\infty \frac{(-1)^n}{4n^2 - 1} = \sum_{n = 1}^\infty \frac{(-1)^n}{(2n - 1)(2n + 1)},$$ noting that $$\frac{1}{2n - 1} = \int_0^1 x^{2n - 2} \, dx \qquad \text{and} \qquad \frac{1}{2n + 1} = \int_0^1 y^{2n} \, dy,$$ the sum can be rewritten as \begin{align*} \sum_{n = 1}^\infty \frac{(-1)^n}{4n^2 - 1} &= \sum_{n = 1}^\infty (-1)^n \int_0^1 \int_0^1 x^{2n - 2} y^{2n} \, dx dy\\ &= \int_0^1 \int_0^1 \sum_{n = 1}^\infty (-1)^n x^{2n - 2} y^{2n} \,dx dy \tag1\\ &= \int_0^1 \int_0^1 \frac{1}{x^2} \sum_{n = 1}^\infty (-x^2 y^2)^n \, dx dy\\ &= \int_0^1 \int_0^1 \frac{1}{x^2} \cdot \frac{-x^2 y^2}{1 + x^2 y^2} \, dx dy \tag2\\ &= - \int_0^1 \int_0^1 \frac{y^2}{1 + x^2 y^2} \, dx dy\\ &= - \int_0^1 \Big{[} y \tan^{-1} (xy) \Big{]}_0^1 \, dy\\ &= -\int_0^1 y \tan^{-1} y \, dy\\ &= -\left [\frac{y^2}{2} \tan^{-1} y \right ]_0^1 + \frac{1}{2} \int_0^1 \frac{y^2}{1 + y^2} \, dy \tag3\\ &= -\frac{\pi}{8} + \frac{1}{2} \int_0^1 \frac{(1 + y^2) - 1}{1 + y^2} \, dy\\ &= -\frac{\pi}{8} + \frac{1}{2} \int_0^1 \left (1 - \frac{1}{1 + y^2} \right ) \, dy\\ &= -\frac{\pi}{8} + \frac{1}{2} \left (1 - \frac{\pi}{4} \right )\\ &= \frac{1}{2} - \frac{\pi}{4}. \end{align*}

Explanation

(1) Interchanging the sum with the double integration.

(2) Summing the series which is geometric.

(3) Integration by parts.

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  • $\begingroup$ Thanks! How do you get rid of the $(-1)^{n}$ at (1)? $\endgroup$ – bencemeszaros Jan 20 '18 at 11:42
  • $\begingroup$ As follows: $(-1)^n x^{2n} y^{2n} = (-1)^n (x^2)^n (y^2)^n = (-x^2y^2)^n$. $\endgroup$ – omegadot Jan 20 '18 at 22:40
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 4n^{2} - 1} & = \sum_{n = 1}^{\infty}{\ic^{2n} \over \pars{2n}^{2} - 1} = \sum_{n = 2}^{\infty}{\ic^{n} \over n^{2} - 1}\,{1 + \pars{-1}^{n} \over 2} = \Re\sum_{n = 2}^{\infty}{\ic^{n} \over n^{2} - 1} \\[5mm] & = {1 \over 2}\,\Re\sum_{n = 2}^{\infty}{\ic^{n} \over n - 1} - {1 \over 2}\,\Re\sum_{n = 2}^{\infty}{\ic^{n} \over n + 1} = {1 \over 2}\,\Re\bracks{\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n}} + {1 \over 2}\,\Re\bracks{\ic\sum_{n = 3}^{\infty}{\ic^{n} \over n}} \\[5mm] & = {1 \over 2}\,\Re\bracks{\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n}} + {1 \over 2} \,\Re\bracks{1 + {\ic \over 2} + \ic\sum_{n = 1}^{\infty}{\ic^{n} \over n}} = {1 \over 2} + \Re\bracks{\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n}} \\[5mm] & = {1 \over 2} - \Im\sum_{n = 1}^{\infty}{\ic^{n} \over n} = {1 \over 2} + \Im\ln\pars{1 - \ic} = \bbx{{1 \over 2} - {\pi \over 4}} \approx -0.2854 \end{align}

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