Sum of Reciprocals of Squares of Odd Integers

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Theorem

\(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) \(=\) \(\displaystyle 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^2} 8\)


Proof 1

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

Hence:

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

$\blacksquare$


Proof 2

Let $n$ be a positive integer.

\(\displaystyle \int_0^1 \frac {x^{2 n + 1} } {\sqrt{1 - x^2} } \mathrm d x\) \(=\) \(\displaystyle \int_0^{\pi / 2} \sin^{2 n + 1} x \ \mathrm d x\) substituting $x \to \sin x$
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {2^{2 n} \left({n!}\right)^2} {\left({2 n + 1}\right)!}\) Reduction Formula for Definite Integral of Power of Sine

We have:

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

We also have:

\(\displaystyle \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \ \mathrm d x\) \(=\) \(\displaystyle \left[{\frac 1 2 \left({\arcsin x}\right)^2}\right]_0^1\) Fundamental Theorem of Calculus
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 8\)

So we can deduce:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\left({2 n + 1}\right)^2} = \frac {\pi^2} 8$


$\blacksquare$


Proof 3

\(\displaystyle \) \(\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2}\) Sum of Reciprocals of Squares of Odd Integers as Double Integral
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \, \mathrm d x \, \mathrm d y\) Sum of Reciprocals of Squares of Odd Integers as Double Integral
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \int_0^1 \frac 1 {\left({1 - x y}\right) \left({1 + x y}\right)} \, \mathrm d x \, \mathrm d y\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \int_0^1 \frac 1 {2 x \left({1 - x y}\right)} \, \mathrm d x \, \mathrm d \left({x y}\right) - \int_0^1 \int_0^1 \frac 1 {2 x \left({1 + x y}\right)} \, \mathrm d x \, \mathrm d \left({x y}\right)\) Primitive of Function of Constant Multiple, partial fraction expansion
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int_0^1 \left[\frac{\ln \left( {1+xy} \right) } x\right]_0^1 \mathrm d x - \frac 1 2\int_0^1 \left[\frac{\ln \left({1 - x y}\right)} x\right]_0^1 \, \mathrm d x\) Primitive of Reciprocal
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int_0^1 \left[{\frac {\ln \left( {1 + x} \right)} x - \frac{\ln 1} x}\right] \mathrm d x - \frac 1 2 \int_0^1 \left[{\frac {\ln \left({1 - x}\right)} x - \frac {\ln 1} x}\right] \, \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int_0^1 \frac 1 x \ln \left({\frac {1 + x} {1 - x} }\right) \, \mathrm d x\) Sum of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \frac{\tanh^{-1} x} x \, \mathrm d x\) Definition of Inverse Hyperbolic Tangent



Proof 4

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


Applying the substitution:

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


the Jacobian determinant is:

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


Under this substitution, the image of the region $\closedint 0 1^2$, that is the unit square, is an isosceles triangle, $\bigtriangleup$ with base/height $\dfrac \pi 2$.

(This may be observed by sketching the region $\set {\tuple {u, v}: u, v \ge 0 \land \map \cos u \ge \map \sin v \land \map \cos v \ge \map \sin u}$.)

Hence,:

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

$\blacksquare$


Proof 5

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

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

Setting $x = \pi$:

\(\displaystyle \size \pi\) \(=\) \(\displaystyle \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 \pi n - \pi} } {\paren {2 n - 1}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {-\pi} } {\paren {2 n - 1}^2}\) Sine and Cosine are Periodic on Reals
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\cos \pi} {\paren {2 n - 1}^2}\) Cosine Function is Even
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {-1} {\paren {2 n - 1}^2}\) Cosine of Multiple of Pi
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac \pi 2\) \(=\) \(\displaystyle \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2n - 1}^2}\) Definition of Absolute Value and rearranging
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 8\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2n - 1}^2}\) multiplying through by $\dfrac \pi 4$

$\blacksquare$


Proof 6

Let $f \left({x}\right)$ be the real function defined on $\left({0 \,.\,.\, 2 \pi}\right)$ as:

$f \left({x}\right) = \begin{cases} -\pi & : 0 < x \le \pi \\ x - \pi & : \pi < x < 2 \pi \end{cases}$

By Fourier Series: $-\pi$ over $\left({0 \,.\,.\, \pi}\right)$, $x - \pi$ over $\left({\pi \,.\,.\, 2 \pi}\right)$, its Fourier series can be expressed as:

$f \left({x}\right) \sim S \left({x}\right) = \displaystyle -\frac \pi 4 + \sum_{n \mathop = 1}^\infty \left({\frac {1 - \left({-1}\right)^n} {n^2 \pi} \cos n x + \frac {\left({-1}\right)^n - 1} n \sin n x}\right)$


Consider the point $x = \pi$.

By Fourier's Theorem:

\(\displaystyle S \left({\pi}\right)\) \(=\) \(\displaystyle \frac 1 2 \left({\lim_{x \mathop \to \pi^+} f \left({x}\right) + \lim_{x \mathop \to \pi^-} f \left({x}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {-\pi + \left({\pi - \pi}\right)} 2\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac \pi 2\)


Thus:

\(\displaystyle -\dfrac \pi 2\) \(=\) \(\displaystyle -\frac \pi 4 + \sum_{n \mathop = 1}^\infty \left({\frac {1 - \left({-1}\right)^n} {n^2 \pi} \cos n \pi + \frac {\left({-1}\right)^n - 1} n \sin n \pi}\right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle -\dfrac \pi 4\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {1 - \left({-1}\right)^n} {n^2 \pi} \cos n \pi\) Sine of Multiple of Pi and simplifying
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac {1 - \left({-1}\right)^{2 r} } {\left({2 r}\right)^2 \pi} \cos 2 r \pi + \sum_{n \mathop = 1}^\infty \frac {1 - \left({-1}\right)^{2 r - 1} } {\left({2 r - 1}\right)^2 \pi} \cos \left({2 r - 1}\right) \pi\) splitting summation into odd and even numbers
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac {1 - 1} {\left({2 r}\right)^2 \pi} \cos 2 r \pi + \sum_{n \mathop = 1}^\infty \frac {1 - \left({-1}\right)} {\left({2 r - 1}\right)^2 \pi} \cos \left({2 r - 1}\right) \pi\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac 2 {\left({2 r - 1}\right)^2 \pi} \cos \left({2 r - 1}\right) \pi\) even terms vanish, and further simplification
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac 2 {\left({2 r - 1}\right)^2 \pi} \left({-1}\right)\) even terms vanish
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\pi^2} 8\) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac 1 {\left({2 r - 1}\right)^2}\) multiplying both sides by $-\dfrac \pi 2$

$\blacksquare$


Proof 7

By Fourier Cosine Series for $x$ over $\left[{0 \,.\,.\, \pi}\right]$:

$\displaystyle x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac{\cos \left({2 n - 1}\right) x} {\left({2 n - 1}\right)^2}$

for $x \in \left[{0 \,.\,.\, \pi}\right]$.


We have that:

$f \left({\pi}\right) = f \left({\pi - 2 \pi}\right) = f \left({-\pi}\right) = \pi$

and so:

$f \left({\pi^-}\right) = f \left({\pi^+}\right)$


Hence we can set $x = \pi$:

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

$\blacksquare$


Also presented as

This result can also be seen presented as:

$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\left({2 n + 1}\right)^2} = \dfrac {\pi^2} 8$


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