Why so Series?

Calculus Level 5

∑ n = 1 ∞ ( − 1 ) n 1 + n 2 \large \sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+n^2} If the series above can be expressed as π a sinh ⁡ ( π ) − b c \dfrac{\pi}{a\sinh(\pi)}-\dfrac{b}{c} , where a a , b b and c c are positive integers. Find the value of a + b + c a+b+c .


The answer is 5.

Hjalmar Orellana Soto by Hjalmar Orellana Soto , 24, Chile

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1 solution

Let f ( x ) = cosh ⁡ ( x ) f(x)=\cosh(x) for x ∈ [ − π , π ] x\in[-\pi,\pi] . The Fourier series of this function in the interval gives us: cosh ⁡ ( x ) = sinh ⁡ ( π ) π − 2 sinh ⁡ ( π ) π ∑ n = 1 ∞ ( − 1 ) n 1 + n 2 cos ⁡ ( n x ) \cosh(x)=\frac{\sinh(\pi)}{\pi}-\frac{2\sinh(\pi)}{\pi}\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{1+n^2} \cos(nx) And evaluating x = 0 x=0 we find: ∑ n = 1 ∞ ( − 1 ) n 1 + n 2 = π 2 sinh ⁡ ( π ) − 1 2 \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{1+n^2} = \frac{\pi}{2\sinh(\pi)}-\frac{1}{2}

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