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Wikipedia's List of formulae involving $\pi$ claims $$\sum_{n=0}^\infty \frac{(-1)^{(n^2-n)/2}}{2n+1}=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\cdots=\frac{\pi}{2\sqrt{2}}$$ and cites Chrystal's Algebra, but I don't have access to that. How can we prove the identity?

It is similar to the famous $$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$ (Why is $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots = \frac{\pi}{4}$?) but the signs are changed. Does it follow from that famous one by some clever series manipulation?

Nomas2
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  • but I don't have access to that --- Many copies of several different editions of Chrystal's "Algebra" are freely available on the internet (google-books, internet archive, and probably other places). Actually, Chrystal wrote TWO different algebra texts, and the one you DON'T want is his "Introduction to Algebra" (first published in 1898). – Dave L. Renfro Nov 01 '23 at 21:29

1 Answers1

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Consider the expansion of $$\frac{1+x^2}{1+x^4}=\sum_{k=0}^\infty(-1)^k(x^{4k}+x^{4k+2})$$ Integrating gives $$\begin{align}\int_0^x\frac{1+t^2}{1+t^4}\,dt&=\sum_{k=0}^\infty(-1)^k\left(\frac{x^{4k+1}}{4k+1}+\frac{x^{4k+3}}{4k+3}\right)\\&=\frac{x}1+\frac{x^3}3-\frac{x^5}5-\frac{x^7}7+\cdots\end{align}$$ Substituting $x=1$ gives $$\int_0^1\frac{1+x^2}{1+x^4}\,dx=\frac11+\frac13-\frac15-\frac17+\cdots$$ This integral comes to $$\begin{align} \int_0^1\frac{1+x^2}{1+x^4}\,dx&=\frac12\int_0^1\frac1{x^2+\sqrt2x+1}+\frac1{x^2-\sqrt2x+1}\,dx\\&=\frac12\int_0^1\frac1{\left(x+\frac{1}{\sqrt2}\right)^2+\frac12}+ \frac1{\left(x-\frac{1}{\sqrt2}\right)^2+\frac12}\,dx\\&=\left[\frac{\arctan\left(\sqrt2x+1\right)+ \arctan\left(\sqrt2x-1\right)}{\sqrt2}\right]_0^1\\&=\frac{\arctan\left(\sqrt2+1\right)+\arctan\left(\sqrt2-1\right)-\arctan(1)-\arctan(-1)}{\sqrt2}\\&=\frac{\frac38\pi+\frac18\pi-\frac14\pi+\frac14\pi}{\sqrt2}\\&=\boxed{\frac\pi{2\sqrt2}}\end{align}$$ as required.

Sai Mehta
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