Sum of Reciprocals of Squares of Odd Integers/Proof 5

Theorem

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)

\(=\)

\(\ds 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots\)

\(\ds \)

\(=\)

\(\ds \dfrac {\pi^2} 8\)

Proof

By Fourier Series of Absolute Value of x, for $x \in \closedint {-\pi} \pi$:

$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {\paren {2 n - 1} x} } {\paren {2 n - 1}^2}$

The validity of the material on this page is questionable. In particular: This formula is proved only for $x \in \openint {-\pi} \pi$. Probably you want to do $x\nearrow \pi$ making use of absolute convergent sum. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Setting $x = \pi$:

\(\ds \size \pi\)

\(=\)

\(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 \pi n - \pi} } {\paren {2 n - 1}^2}\)

\(\ds \)

\(=\)

\(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {-\pi} } {\paren {2 n - 1}^2}\)

Sine and Cosine are Periodic on Reals

\(\ds \)

\(=\)

\(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\cos \pi} {\paren {2 n - 1}^2}\)

Cosine Function is Even

\(\ds \)

\(=\)

\(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {-1} {\paren {2 n - 1}^2}\)

Cosine of Multiple of Pi

\(\ds \leadsto \ \ \)

\(\ds \frac \pi 2\)

\(=\)

\(\ds \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)

Definition of Absolute Value and rearranging

\(\ds \leadsto \ \ \)

\(\ds \frac {\pi^2} 8\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)

multiplying through by $\dfrac \pi 4$

$\blacksquare$