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How does one prove the given series? $$\sum_{n=0}^\infty\left(\frac{(-1)^n}{4n-a+2}+\frac{(-1)^n}{4n+a+2}\right)=\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right)$$

This series came up in xpaul's calculation during the process of answering my homework problem. I really appreciate his help for me but I am looking a method to prove the above series using a real analysis method because the link he cited (Ron G's answer) to help me to prove it is using the residue theorem. I worked for a while on this today but was unsuccessful. $$\begin{align}\sum_{n=0}^\infty\left(\frac{(-1)^n}{4n-a+2}+\frac{(-1)^n}{4n+a+2}\right)&=4\sum_{n=0}^\infty(-1)^n\frac{2n+1}{(4n+2)^2-a^2}\\&=\sum_{n=0}^\infty(-1)^n\frac{2n+1}{(2n+1)^2-\left(\frac{a}{2}\right)^2}\end{align}$$ Comparing with Taylor series for secant, hyperbolic secant, or any other well known series forms but I could not get any of them to work, perhaps someone else can? I would like a nice proof and avoiding residue method in order to complete my homework's answer. Would you help me? Any help would be appreciated. Thanks in advance.

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Venus
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2 Answers2

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$$\begin{eqnarray*}\sum_{n=0}^\infty\left(\frac{(-1)^n}{4n-a+2}+\frac{(-1)^n}{4n+a+2}\right)&=&\sum_{n=0}^{+\infty}\int_{0}^{1}(-1)^n x^{4n+1}\left(x^a+x^{-a}\right)\,dx \\&=& \int_{0}^{1}\frac{x}{1+x^4}\left(x^a+x^{-a}\right)\,dx\end{eqnarray*}$$ hence we have that our integral equals: $$ \frac{1}{8}\left(\psi\left(\frac{6+a}{8}\right)-\psi\left(\frac{2+a}{8}\right)+\psi\left(\frac{6-a}{8}\right)-\psi\left(\frac{2-a}{8}\right)\right)$$ but since: $$ \psi(x)-\psi(1-x) = -\pi\cot(\pi x) $$ the previous line equals: $$ \frac{\pi}{8}\left(\tan\left(\frac{\pi}{8}(a+2)\right)-\tan\left(\frac{\pi}{8}(a-2)\right)\right) $$ that simplifies to: $$ \frac{\pi}{4}\sec\frac{a\pi}{4}$$ as wanted.

Jack D'Aurizio
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  • What is $\psi$? – TonyK Nov 13 '14 at 11:25
  • @TonyK: the logarithmic derivatives of the $\Gamma$ function: $$\psi(x)=\frac{\Gamma'}{\Gamma}(x).$$ See also http://en.wikipedia.org/wiki/Digamma_function – Jack D'Aurizio Nov 13 '14 at 11:30
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    Thanks Jack, I'll take this answer into consideration. +1 – Venus Nov 13 '14 at 11:46
  • Are there others forms for the other common functions, I mean sin(x), cos(x), and ln(x) – E.H.E Nov 13 '14 at 11:47
  • Technically, that integral has a closed form for all rational $a$—so you can prove it that way—but it's probably messy. – Akiva Weinberger Nov 13 '14 at 11:55
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    Aren't you just sweeping the residue theorem stuff under the carpet here? Or can that integral be evaluated using real-analytic methods? – TonyK Nov 13 '14 at 12:12
  • @TonyK: I am just exploiting the properties of Euler Beta function (its logarithmic derivatives, for real) and the reflection formula for the $\Gamma$ function, no residues are involved (or hidden under the carpet) in this particular proof. – Jack D'Aurizio Nov 13 '14 at 12:17
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I answered this before in Closed form of $\int_{0}^{\infty} \frac{\tanh(x)\,\tanh(2x)}{x^2}\;dx$. However the solution was too long as Venus mentioned. Inspired from Jack D'aurizio's answer, I have a simple solution for this. It is easy to check that \begin{eqnarray*} &&\sum_{n=0}^\infty\left(\frac{(-1)^n}{4n-a+2}+\frac{(-1)^n}{4n+a+2}\right)\\ &=& \int_{0}^{1}\frac{x}{1+x^4}\left(x^a+x^{-a}\right)\,dx=\int_{0}^{\infty}\frac{x^{a+1}}{1+x^4}\,dx\\ &=&\frac{1}{a+2}\int_0^\infty\frac{1}{1+x^{\frac{a+2}{4}}}dx. \end{eqnarray*} Now using the following well-known integral $$ \int_0^\infty\frac{1}{1+x^n}dx=\frac{\pi}{n\sin(\pi/n)}, \text{ for }n>1, $$ (for example, see Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only) it is easy to get the answer.

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xpaul
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  • Wait a sec, I'm a bit lost for this step: $$\int_{0}^{1}\frac{x}{1+x^4}\left(x^a+x^{-a}\right),dx=\int_{0}^{\infty}\frac{x^{a+1}}{1+x^4},dx$$ What did you do? I only thought about substitution $x=\frac{1}{u}$ but didn't arrive to your result. Can you elaborate? – Venus Nov 13 '14 at 15:46
  • OK. I get it now. Many thanks for your nice answer & your help (again!) +1 – Venus Nov 13 '14 at 15:56