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Looking to the following series:

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

It converges according to Leibniz criteria. However it does not seem to be a telescopic series (if you take partial fractions, you end up with two positive terms), neither a geometric one, so I can not figure out a way to find its sum. Maybe I am missing something here.

Thanks for your time and I appreciate any help.

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5 Answers 5

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It is in fact telescopic. For the partial sum: $$ S_N = \sum_{n=1}^N (-1)^n \left( \frac{1}{2n-1}+\frac{1}{2n+1}\right)= \sum_{n=1}^N \frac{(-1)^n}{2n-1} - \sum_{k=2}^{N+1} \frac{(-1)^k}{2k-1} =-1+\frac{(-1)^N}{2N+1}\rightarrow -1$$

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It is the moment for the extreme overkill. By differentiation under the integral sign we have that $$ \sum_{n\geq 1}\frac{4n\cos(\pi n x)}{4n^2-1}\tag{1} $$ over the interval $x\in(0,2)$ is the Fourier cosine series of $$ f(x) = -1+\cos\left(\frac{\pi x}{2}\right)\cdot\log\cot\left(\frac{\pi x}{4}\right)\tag{2} $$ hence by evaluating $(2)$ at $x=1$ $$ \sum_{n\geq 1}\frac{4n(-1)^n}{4n^2-1}=\color{red}{-1}\tag{3} $$ follows.

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Partial fractions works:

$$\frac{(-1)^n4n}{4n^2-1}=\frac{(-1)^{n}}{2n+1}-\frac{(-1)^{n-1}}{2n-1}$$

This is a telescoping sum of the form:

$$\sum_{n=1}^{x} \left(a_{n+1}-a_{n}\right)=a_{x+1}-a_1$$

With:

$$a_{n}=\frac{(-1)^{n-1}}{2n-1}$$

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This is another "extreme overkill."

Recalling that the Taylor Series for the arctangent function is given by

$$\arctan(x)=\sum_{n=0}\frac{(-1)^{n}x^{2n+1}}{2n+1}$$

we find that

$$\begin{align} \sum_{n=1}^\infty \frac{4n(-1)^n}{4n^2-1}&=\sum_{n=1}^\infty \left(\frac{(-1)^{n}}{2n+1}+\frac{(-1)^{n}}{2n-1}\right)\\\\ &=\left(\arctan(1)-1\right)-\arctan(1)\\\\ &=-1 \end{align}$$

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    $\begingroup$ You didn't actually need to know the relation to $\arctan 1$ to make this substitution. And if you keep the sums as sums, it's the same as the telescoping proof. $\endgroup$ Aug 27, 2016 at 5:11
  • $\begingroup$ @MarioCarneiro To what "substitution" are you referring? There was none. And I admonish you to actually read the preamble before making such an "out-of-context" comment. ;-)) $\endgroup$
    – Mark Viola
    Aug 27, 2016 at 15:32
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Third overkill. We have $$\sum_{n\geq1}\frac{\left(-1\right)^{n}4n}{4n^{2}-1}=2\sum_{n\geq1}\frac{\left(-1\right)^{n}2n-1+1}{4n^{2}-1} $$ $$=2\sum_{n\geq1}\frac{\left(-1\right)^{n}}{2n+1}+2\sum_{n\geq1}\frac{\left(-1\right)^{n}}{4n^{2}-1} $$ $$=\frac{\pi}{2}-2+1-\frac{\pi}{2}=\color{red}{-1} $$ where the first sum follows from the Taylor series of the $\arctan $ function and the second follows from the summation formula $$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \textrm{residues of }\pi\csc\left(\pi z\right)f(z)\textrm{ at }f\left(z\right)\textrm{ poles}\right\} $$ observing that $$2\sum_{n\geq1}\frac{\left(-1\right)^{n}}{4n^{2}-1}=\sum_{n\in\mathbb{Z}}\frac{\left(-1\right)^{n}}{4n^{2}-1}+1.$$

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