Expert Answer to "Proving:∑n=0∞sin4(4n+2)(2n+1)2=5π216−3π4"

Cobie Sadler

Answered question

2022-02-24

Proving:

\sum_{n = 0}^{\infty} \frac{\sin^{4} (4 n + 2)}{{(2 n + 1)}^{2}} = \frac{5 π^{2}}{16} - \frac{3 π}{4}

publilandiaik8

Beginner2022-02-25Added 5 answers

Since:

\sin x = \frac{1}{2 i} (e^{i x} - e^{- i x})

we have:

\sin^{4} x = \frac{1}{16} (e^{4 i x} + e^{- 4 i x} - 4 e^{2 i x} - 4 e^{2 i x} + 6) = \frac{1}{8} (\cos (4 x) - 4 \cos (2 x) + 3)

It follows:

S = \sum_{n \geq 0} \frac{\sin^{4} (4 n + 2)}{{(2 n + 1)}^{2}}

= \frac{1}{8} \sum_{n \geq 0} \frac{\cos (16 n + 8)}{{(2 n + 1)}^{2}} - \frac{1}{2} \sum_{n \geq 0} \frac{\cos (8 n + 4)}{{(2 n + 1)}^{2}} + \frac{3}{8} \sum_{n \geq 0} \frac{1}{{(2 n + 1)}^{2}}

where

\sum_{n \geq 0} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8}

. Now it happens that:

f (x) = \sum_{n \geq 0} \frac{\cos ((2 n + 1) x)}{{(2 m + 1)}^{2}}

is the Fourier series of the triangle wave with period

2 π

and amplitude

2 \cdot \frac{π^{2}}{8}

, hence:

f (4) = π - 3 \frac{π^{2}}{8}, f (2) = - \frac{π}{2} + \frac{π^{2}}{8}

and:

S = \frac{1}{8} - \frac{1}{2} f (2) + \frac{3}{8} f (0) = \frac{5 π^{2}}{16} - \frac{3 π}{4}

as wanted.

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