Sum of Reciprocals of Squares of Odd Integers/Proof 5
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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 $\dfrac \pi 4$ |
$\blacksquare$