An Alternating Sum

Calculus Level 5

S=∑n=−∞∞(−1)n1+n2=aπebπecπ−1,S=\sum_{n=-\infty}^{\infty} \dfrac{(-1)^n}{1+n^2}=\dfrac{a\pi e^{b\pi}}{e^{c\pi}-1},

where a,b,ca,b,c are natural numbers. Find a+b+ca+b+c.

Note: Please avoid using wolfram alpha.

Hint: Think about the Fourier transform of f(x)=e−∣ax∣f(x)=e^{-|ax|}.

Pratik Shastri by Pratik Shastri
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