Prove that : ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) 3 = π 3 32 .

Question

Prove that :      ∑          k      =      0              ∞                  (      −      1              )                  k                            (      2      k      +      1              )        3              =            π      3        32    .

Nkgopotsev1g · Accepted Answer

You can evaluate this sum using the residue theorem. First, note that it may be extended out to

- \infty

:

\sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{(2 k + 1)^{3}} = \frac{1}{2} \sum_{k = - \infty}^{\infty} \frac{(- 1)^{k}}{(2 k + 1)^{3}}

From the residue theorem:

\sum_{k = - \infty}^{\infty} \frac{(- 1)^{k}}{(2 k + 1)^{3}} = - {Res}_{z = - 1 / 2} \frac{π \csc π z}{(2 z + 1)^{3}} = - \frac{1}{8} {Res}_{z = - 1 / 2} \frac{π \csc π z}{(z + 1 / 2)^{3}}

This residue involves taking the second derivative of the csc term. Note that, for a generic function f(z) having a triple pole at

z = z_{0}

:

{Res}_{z = z_{0}} f (z) = \frac{1}{2!} lim_{z \to z_{0}} \frac{d^{2}}{d z^{2}} [(z - z_{0})^{3} f (z)]

so that

{Res}_{z = - 1 / 2} \frac{π \csc π z}{(z + 1 / 2)^{3}} = \frac{π^{3}}{2!} {[\csc π z \cot^{2} π z + \csc^{3} π z]}_{z = - 1 / 2} = - \frac{π^{3}}{2}

Putting this all together, we get the stated result:

\sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{(2 k + 1)^{3}} = \frac{π^{3}}{32}

Aryanna Fisher · Accepted Answer

Here is another way using Fourier analysis: Let

f (t) = {\begin{cases} t - t^{2} & 0 < t < 1 \\ - f (- t) & - 1 < t < 0 \end{cases}

be a function with period 2. Then we can express f in a Fourer series:

f (t) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos n π t + b_{n} \sin n π t

where

\begin{aligned} a_{n} = \frac{1}{1} \int_{0}^{2} f (t) \cos n π t d t \\ b_{n} = \frac{1}{1} \int_{0}^{2} f (t) \sin n π t d t \end{aligned}

But f is odd, so

a_{n} = 0

. It follows that

\begin{aligned} b_{n} & = \int_{0}^{2} f (t) \sin n π t d t = {f (t) \sin n π t even} = 2 \int_{0}^{1} f (t) \sin n π t d t = \\ = 2 \int_{0}^{1} (t - t^{2}) \sin n π t d t = \frac{4 - 4 (- 1)^{n}}{n^{3} π^{3}} \end{aligned}

Plugging in

t = \frac{1}{2}

, we get

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

But

1 - (- 1)^{n} = 0

only if n is even, so

\frac{1}{16} = \sum_{k = 0}^{\infty} \frac{2 (- 1)^{k}}{(2 k + 1)^{3} π^{3}} ⟺ \sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{(2 k + 1)^{3}} = \frac{π^{3}}{32}

Prove that : sum_(k=0)^infty (-1)^k/(2k+1)^3=pi^3/32

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