# Sum of Reciprocals of Squares of Odd Integers

## 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:

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

the Jacobian matrix is:

$\mathbf J_{\mathbf f} := \begin{pmatrix}  {\dfrac \partial {\partial u} \dfrac {\map \sin u} {\map \cos v} } & {\dfrac \partial {\partial v} \dfrac {\map \sin u} {\map \cos v} } \\ {\dfrac \partial {\partial u} \dfrac {\map \sin v} {\map \cos u} } & {\dfrac \partial {\partial v} \dfrac {\map \sin v} {\map \cos u} }  \end{pmatrix}$

and the Jacobian determinant is:

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

Under this substitution, the image of the region $\closedint 0 1^2$, that is the unit square, is an isosceles triangle, $\bigtriangleup$ with base and height $\dfrac \pi 2$: Vertices: $\tuple {0, 0}; \tuple {0, \dfrac \pi 2}; \tuple {\dfrac \pi 2, 0}$.

We have:

$0 \le \dfrac {\map \sin u} {\map \cos v} \le 1$
$\leadsto u \ge 0$ and
$\leadsto \map \sin u \le \map \cos v$

And we have:

$0 \le \dfrac {\map \sin v} {\map \cos u} \le 1$
$\leadsto v \ge 0$ and
$\leadsto \map \sin v \le \map \cos u$

This gives us the region: $\set {\tuple {u, v}: u, v \ge 0 \land \map \cos u \ge \map \sin v \land \map \cos v \ge \map \sin u}$.

Which is equivalent to the region: $\set {\tuple {u, v}: u, v \ge 0 \land v \le \dfrac \pi 2 - u }$.

Hence,:

 $\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2y^2} \rd x \rd y$ $=$ $\ds \iint_{\bigtriangleup} \frac {1 - \map {\tan^2} u \, \map {\tan^2} v} {1 - \paren {\map \tan u \, \map \tan v}^2} \rd u \rd v$ $\ds$ $=$ $\ds \iint_{\bigtriangleup} \rd u \rd v$ $\ds$ $=$ $\ds \frac 1 2 \int_0^{\frac \pi 2} \int_0^{\frac \pi 2} \rd u \rd v$ 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 $x \in \closedint {-\pi} \pi$:

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

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 $x = \pi$.

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

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

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

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:

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