Infinite sums can be manipulated!

Algebra Level 3

$\large \sum_{n=1}^\infty \frac 1{(2n-1)^2}$

Given that $\zeta (2) = \dfrac {\pi^2}6$ , find the value of the infinite sum above.

Notation: $\displaystyle \zeta (s) = \sum_{n=1}^\infty \frac 1{n^s}$ for $s \in \mathbb C \ \forall \ \Re (s) > 1$ denotes the Riemann zeta function .

\frac { { \pi }^{ 2 } }{ 2 }

1

\frac { { \pi }^{ 2 } }{ 3 }

\frac { { \pi }^{ 2 } }{ 8 }

1 solution

Chew-Seong Cheong
Sep 2, 2016

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{(2n-1)^2} \\ & = 1 + \frac 1{3^2} + \frac 1{5^2} + \frac 1{7^2} + \frac 1{9^2} + ... \\ & = 1 + \color{#3D99F6}{\frac 1{2^2}} + \frac 1{3^2} + \color{#3D99F6}{\frac 1{4^2}} + \frac 1{5^2} + \color{#3D99F6}{\frac 1{6^2}} + \frac 1{7^2} + ... \color{#D61F06}{-} \left(\color{#3D99F6}{\frac 1{2^2}} + \color{#3D99F6}{\frac 1{4^2}} + \color{#3D99F6}{\frac 1{6^2}} + \color{#3D99F6}{\frac 1{8^2}} + ... \right) \\ & = 1 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \frac 1{5^2} + ... - \frac 1{2^2} \left(1 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \frac 1{5^2} + ... \right) \\ & = \sum_{n=1}^\infty \frac 1{n^2} - \frac 14 \sum_{n=1}^\infty \frac 1{n^2} \\ & = \frac 34 \zeta (2) = \frac 34 \times \frac {\pi^2}6 = \boxed{\dfrac {\pi^2}8} \end{aligned}$

Tapas Mazumdar , I have changed the wording of your problem. You should use references in Brilliant if available instead of Wikipedia or other external ones. You can do a search in the search box on top.

Chew-Seong Cheong - 4 years, 9 months ago

A great solution to the problem. P.S. Thanks for the word of advice. 😊

Tapas Mazumdar - 4 years, 9 months ago

Infinite sums can be manipulated!

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