Sum of Reciprocals of Squares Alternating in Sign/Proof 3

Theorem

\(\ds \dfrac {\pi^2} {12}\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \frac 1 {1^2} - \frac 1 {2^2} + \frac 1 {3^2} - \frac 1 {4^2} + \cdots\)

Proof

Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:

$\map f x = \pi^2 - x^2$

By Fourier Series: $\pi^2 - x^2$ over $\openint {-\pi} \pi$:

$\ds \pi^2 - x^2 \sim \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}$

for $x \in \openint {-\pi} \pi$.

Setting $x = 0$:

\(\ds \pi^2 - 0^2\)

\(=\)

\(\ds \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos 0} {n^2}\)

\(\ds \)

\(=\)

\(\ds \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac 1 {n^2}\)

Cosine of Zero is One

\(\ds \leadsto \ \ \)

\(\ds \frac {\pi^2} {12}\)

\(=\)

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

simplification

\(\ds \)

\(=\)

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

rearrangement

$\blacksquare$

Sources

• 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $6$.