Sum of Reciprocals of Squares of Odd Integers

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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 1

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\ds \map \zeta 2\) Definition of Riemann Zeta Function
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^2} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)
\(\ds \) \(=\) \(\ds \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)
\(\ds \) \(=\) \(\ds \frac 1 4 \map \zeta 2 + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)

Hence:

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) \(=\) \(\ds \frac 3 4 \map \zeta 2\)
\(\ds \) \(=\) \(\ds \frac 3 4 \cdot \frac{\pi^2} 6\) Basel Problem
\(\ds \) \(=\) \(\ds \frac {\pi^2} 8\)

$\blacksquare$


Proof 2

Let $n$ be a positive integer.

\(\ds \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \rd x\) \(=\) \(\ds \int_0^{\pi / 2} \sin^{2 n + 1} x \rd x\) substituting $x \to \sin x$
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac {2^{2 n} \paren {n!}^2} {\paren {2 n + 1}!}\) Reduction Formula for Definite Integral of Power of Sine

We have:

\(\ds \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \rd x\) \(=\) \(\ds \int_0^1 \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \rd x\) Power Series Expansion for Real Arcsine Function
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \rd x\) Interchange of sum and integral is valid by Tonelli's Theorem
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } \frac {2^{2 n} \paren {n!}^2} {\paren {2 n + 1}!}\) by $(1)$
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2}\)

We also have:

\(\ds \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \rd x\) \(=\) \(\ds \intlimits {\frac 1 2 \paren {\arcsin x}^2} 0 1\) Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds \frac {\pi^2} 8\)

So we can deduce:

$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} = \frac {\pi^2} 8$


$\blacksquare$


Proof 3

This "proof" leads to a dead end. Its value lies in showing that this approach does not work.

\(\ds \) \(\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)
\(\ds \) \(=\) \(\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y\) Sum of Reciprocals of Squares of Odd Integers as Double Integral
\(\ds \) \(=\) \(\ds \int_0^1 \int_0^1 \frac 1 {\paren {1 + x y} \paren {1 - x y} } \rd x \rd y\)
\(\ds \) \(=\) \(\ds \int_0^1 \int_0^1 \frac 1 {2 \paren {1 + x y} } \rd x \rd y - \int_0^1 \int_0^1 \frac 1 {2 \paren {1 - x y} } \rd x \rd y\) Primitive of Function of Constant Multiple, partial fraction expansion
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^1 \intlimits {\frac {\map \ln {1 + x y} } x} 0 1 \rd x - \frac 1 2 \int_0^1 \intlimits {\frac {\map \ln {1 - x y} } x} 0 1 \rd x\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^1 \paren {\frac {\map \ln {1 + x} } x - \frac {\ln 1} x} \rd x - \frac 1 2 \int_0^1 \paren {\frac {\map \ln {1 - x} } x - \frac {\ln 1} x} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^1 \frac 1 x \map \ln {\frac {1 + x} {1 - x} } \rd x\) Sum of Logarithms
\(\ds \) \(=\) \(\ds \int_0^1 \frac {\artanh x} x \rd x\) Definition of Real Area Hyperbolic Tangent
\(\ds \) \(=\) \(\ds \int_0^1 \frac 1 x \paren {x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots} \rd x\) Power Series Expansion for Real Area Hyperbolic Tangent
\(\ds \) \(=\) \(\ds \int_0^1 \paren {1 + \frac {x^2} 3 + \frac {x^4} 5 + \frac {x^6} 7 + \cdots} \rd x\)
\(\ds \) \(=\) \(\ds \intlimits {x + \frac {x^3} 9 + \frac {x^5} {25} + \frac {x^7} {49} + \cdots} 0 1\)
\(\ds \) \(=\) \(\ds 1 + \frac 1 9 + \frac 1 {25} + \frac 1 {49} + \cdots\) and a dead end is reached


Proof 4

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2n - 1}^2}\) \(=\) \(\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2y^2} \rd x \rd y\) Sum of Reciprocals of Squares of Odd Integers as Double Integral


Applying the substitution:

$\tuple {x, y} = \tuple {\dfrac {\sin u} {\cos v}, \dfrac {\sin v} {\cos u} }$

the Jacobian matrix is:

$\mathbf J_{\mathbf f} := \begin{pmatrix}

{\dfrac \partial {\partial u} \dfrac {\sin u} {\cos v} } & {\dfrac \partial {\partial v} \dfrac {\sin u} {\cos v} } \\ {\dfrac \partial {\partial u} \dfrac {\sin v} {\cos u} } & {\dfrac \partial {\partial v} \dfrac {\sin v} {\cos u} } \end{pmatrix}$

and the Jacobian determinant is:

\(\ds \size {\frac {\partial \tuple {x, y} } {\partial \tuple {u, v} } }\) \(=\) \(\ds \frac \partial {\partial u} \frac {\sin u} {\cos v} \frac \partial {\partial v} \frac {\sin v} {\cos u} - \frac \partial {\partial v} \frac {\sin u} {\cos v} \frac \partial {\partial u} \frac {\sin v} {\cos u}\)
\(\ds \) \(=\) \(\ds \frac {\cos u \, \cos v} {\cos v \, \cos u} - \frac {\sin^2 u \, \sin^2 v} {\cos^2 v \, \cos^2 u}\) Derivative of Secant Function, Derivative of Sine Function
\(\ds \) \(=\) \(\ds 1 - \tan^2 u \tan^2 v\)


Under this substitution, the image of the region $\closedint 0 1^2$, that is the unit square, is an isosceles triangle $\bigtriangleup$ with:

base and height $\dfrac \pi 2$
vertices: $\tuple {0, 0}; \tuple {0, \dfrac \pi 2}; \tuple {\dfrac \pi 2, 0}$


We have:

\(\ds 0\) \(\le\) \(\ds \dfrac {\sin u} {\cos v}\) \(\ds \le 1\)
\(\ds \leadsto \ \ \) \(\ds u\) \(\ge\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sin u\) \(\le\) \(\ds \cos v\)

and we have:

\(\ds 0\) \(\le\) \(\ds \dfrac {\sin v} {\cos u}\) \(\ds \le 1\)
\(\ds \leadsto \ \ \) \(\ds v\) \(\ge\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sin v\) \(\le\) \(\ds \cos u\)


This gives us the region:

$\set {\tuple {u, v}: u, v \ge 0 \land \cos u \ge \sin v \land \cos v \ge \sin u}$

which is equivalent to the region:

$\set {\tuple {u, v}: u, v \ge 0 \land v \le \dfrac \pi 2 - u}$


By Change of Variables Theorem (Multivariable Calculus):

\(\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y\) \(=\) \(\ds \iint_{\bigtriangleup} \frac {1 - \tan^2 u \, \tan^2 v} {1 - \paren {\tan u \, \tan v}^2} \rd u \rd v\)
\(\ds \) \(=\) \(\ds \iint_{\bigtriangleup} \rd u \rd v\)
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^{\frac \pi 2} \int_0^{\frac \pi 2} \rd u \rd v\) Area of Triangle in Terms of Side and Altitude
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac \pi 2}^2\) Primitive of Constant
\(\ds \) \(=\) \(\ds \frac {\pi^2} 8\)

$\blacksquare$


Proof 5

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.
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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 $\dfrac \pi 4$

$\blacksquare$


Proof 6

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

$\map f x = \begin{cases}

-\pi & : 0 < x \le \pi \\ x - \pi & : \pi < x < 2 \pi \end{cases}$

By Fourier Series: $-\pi$ over $\openint 0 \pi$, $x - \pi$ over $\openint \pi {2 \pi}$, its Fourier series can be expressed as:

$\map f x \sim \map S x = \displaystyle -\dfrac \pi 4 + \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {\cos \paren {2 r + 1} x} {\paren {2 r + 1}^2} - \sum_{n \mathop = 1}^\infty \dfrac {2 - \paren {-1}^n \sin n x} n$


Consider the point $x = \pi$.

By Fourier's Theorem:

\(\ds \map S \pi\) \(=\) \(\ds \frac 1 2 \paren {\lim_{x \mathop \to \pi^+} \map f x + \lim_{x \mathop \to \pi^-} \map f x}\)
\(\ds \) \(=\) \(\ds \dfrac {-\pi + \paren {\pi - \pi} } 2\)
\(\ds \) \(=\) \(\ds -\dfrac \pi 2\)


Thus:

\(\ds -\dfrac \pi 2\) \(=\) \(\ds -\dfrac \pi 4 + \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {\cos \paren {2 r + 1} \pi} {\paren {2 r + 1}^2} - \sum_{n \mathop = 1}^\infty \dfrac {2 - \paren {-1}^n \sin n \pi} n\)
\(\ds \leadsto \ \ \) \(\ds -\dfrac \pi 4\) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {\cos \paren {2 r + 1} \pi} {\paren {2 r + 1}^2}\) Sine of Multiple of Pi and simplifying
\(\ds \) \(=\) \(\ds \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {-1} {\paren {2 r + 1}^2}\) Cosine of Multiple of Pi and simplifying
\(\ds \leadsto \ \ \) \(\ds \dfrac {\pi^2} 8\) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \frac 1 {\paren {2 r - 1}^2}\) multiplying both sides by $-\dfrac \pi 2$ and adjusting indices

$\blacksquare$


Proof 7

By Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$:

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

for $x \in \openint 0 \pi$.


We have that:

$\map f \pi = \map f {\pi - 2 \pi} = \map f {-\pi} = \pi$

and so:

$\map f {\pi^-} = \map f {\pi^+}$


Hence we can set $x = \pi$:

\(\ds \pi\) \(=\) \(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} \pi} {\paren {2 n - 1}^2}\)
\(\ds \) \(=\) \(\ds \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {-1} {\paren {2 n - 1}^2}\) Cosine of Multiple of Pi
\(\ds \leadsto \ \ \) \(\ds \frac {\pi^2} 8\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) simplification

$\blacksquare$


Also presented as

This result can also be seen presented as:

$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} = \dfrac {\pi^2} 8$


Sources

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