Sum of Reciprocals of Squares Alternating in Sign

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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 1

Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:

$\map f x = \begin {cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$

From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:

$(1): \quad \map f x \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$

We have that:

\(\ds \map f {\pi - 0}\) \(=\) \(\ds \paren {\pi - \pi}^2\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \map f {\pi + 0}\) \(=\) \(\ds \pi^2\)

where $\map f {\pi - 0}$ and $\map f {\pi + 0}$ denote the limit from the left and limit from the right respectively of $\map f \pi$.


It is apparent that $\map f x$ satisfies the Dirichlet conditions:

$(\text D 1): \quad f$ is bounded on $\openint 0 {2 \pi}$
$(\text D 2): \quad f$ has a finite number of local maxima and local minima.
$(\text D 3): \quad f$ has $1$ discontinuity, which is finite.


Hence from Fourier's Theorem:

\(\ds \map f \pi\) \(=\) \(\ds \frac {\map f {\pi - 0} + \map f {\pi + 0} } 2\)
\(\ds \) \(=\) \(\ds \frac {0 + \pi^2} 2\)
\(\ds \) \(=\) \(\ds \frac {\pi^2} 2\)


Thus setting $x = \pi$ in $(1)$:

\(\ds \map f \pi\) \(=\) \(\ds \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n \pi} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n \pi}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\pi^2} 2\) \(=\) \(\ds \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac {\cos n \pi} {n^2}\) Sine of Multiple of Pi
\(\ds \leadsto \ \ \) \(\ds \frac {\pi^2} 4\) \(=\) \(\ds \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {n^2}\) Cosine of Multiple of Pi and simplification
\(\ds \leadsto \ \ \) \(\ds -\frac {\pi^2} {12}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {n^2}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\pi^2} {12}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } {n^2}\) changing sign and subsuming into powers of $-1$

$\blacksquare$


Proof 2

\(\ds \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } {n^2}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2} - \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n}\right)^2}\) separating odd positive terms from even negative terms
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2} - \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\ds \) \(=\) \(\ds \frac {\pi^2} 8 - \frac 1 4 \times \frac {\pi^2} 6\) Sum of Reciprocals of Squares of Odd Integers, Basel Problem
\(\ds \) \(=\) \(\ds \frac {\pi^2 \left({3 - 1}\right)} {24}\)
\(\ds \) \(=\) \(\ds \frac {\pi^2} {12}\)

$\blacksquare$


Proof 3

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

$\blacksquare$


Also see


Sources

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