Basel Problem

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Theorem

$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.


Proof 1

By Riemann Zeta Function as a Multiple Integral,

$\displaystyle \zeta \left( {2} \right) = \int_0^1 \int_0^1 \frac 1 {1 - xy} \mathrm d A$

Let $\displaystyle \left({u, v}\right) = \left({\frac{x + y} 2, \frac{y - x} 2}\right)$ so that:

$\left({x, y}\right) = \left({u - v, \ u + v}\right)$

Let:

$\left|{J}\right| = \left\vert{\dfrac{\partial \left({x, y}\right)} {\partial \left({u, v}\right)} }\right\vert = 2$

Then, by Change of Variables Theorem (Multivariable Calculus):

$\displaystyle \zeta \left({2}\right) = 2 \iint \limits_S \frac{\mathrm d u \ \mathrm d v} {1 - u^2 + v^2}$

where $S$ is the square defined by the coordinates:

$\left({0, 0}\right), \ \left({\dfrac 1 2, -\dfrac 1 2}\right), \ \left({1, 0}\right), \ \left({\dfrac 1 2, \dfrac 1 2}\right)$

Exploiting the symmetry of the square and the function over the $u$-axis, we have:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \! \int_0^u \frac {\mathrm d v \ \mathrm d u} {1 - u^2 + v^2} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac {\mathrm d v \ \mathrm d u} {1 - u^2 + v^2}}\right)$

Factoring $1-u^2$ gives us:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \! \int_0^u \frac 1 {1 - u^2} \ \frac {\mathrm d v \ \mathrm d u} {\frac {v^2} {1 - u^2} + 1} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac 1 {1 - u^2} \frac {\mathrm d v \ \mathrm d u} {\frac {v^2} {1 - u^2} + 1}}\right)$

and letting:

$s = \dfrac v {\sqrt{1 - u^2} }, \mathrm d s = \dfrac 1 {\sqrt{1 - u^2} }$

allows us to make a substitution into each integral, giving:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \frac 1 {\sqrt{1 - u^2} } \arctan \left({\frac u {\sqrt{1 - u^2} } }\right) \mathrm d u + \int_{\frac 1 2}^1 \frac 1 {\sqrt{1 - u^2} } \arctan \left({\frac{1 - u} {\sqrt{1 - u^2} } }\right) \mathrm d u}\right)$

Consider the right triangle with sides $1$, $x$ and $\sqrt{1 - x^2}$.

From it arises the identity:

$\arcsin x = \arctan \dfrac x {\sqrt{1 - x^2} }$

Let:

\(\displaystyle \theta\) \(=\) \(\displaystyle \arctan \left({\dfrac{1 - u} {\sqrt{1 - u^2} } }\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle \tan^2 \theta\) \(=\) \(\displaystyle \frac {1 - u} {1 + u}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \sec^2 \theta\) \(=\) \(\displaystyle \frac 2 {1 + u}\) Difference of Squares of Secant and Tangent
\(\displaystyle \implies \ \ \) \(\displaystyle \cos 2 \theta\) \(=\) \(\displaystyle u\) Double Angle Formula for Cosine: Corollary 1
\(\displaystyle \implies \ \ \) \(\displaystyle \theta\) \(=\) \(\displaystyle \dfrac 1 2 \arccos u\) taking arccosines of both sides
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 4 - \frac {\arcsin u} 2\) Sine of Complement equals Cosine


This allows us to convert the arctangents from the integrals into arcsines:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} {\frac {\arcsin u} {\sqrt{1 - u^2} }\, \mathrm d u} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt{1 - u^2} } \left({\frac \pi 4 - \frac {\arcsin u} 2}\right)\,\mathrm d u} }\right)$

Substituting:

$s = \arcsin u$, $\mathrm d s = \dfrac 1 {\sqrt{1 - u^2} }$

into the arcsines, and splitting the second integral:

$\zeta \left({2}\right) = 4 \left({\dfrac {\pi^2} {72} + \dfrac {\pi^2} {36} }\right) = \dfrac {\pi^2} 6$

$\blacksquare$


Proof 2

Let:

$\displaystyle P_k = x \prod_{n \mathop = 1}^k \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$

We note that:

\(\displaystyle P_k - P_{k - 1}\) \(=\) \(\displaystyle \left({- \frac {x^3} {k^2 \pi^2} }\right) \prod_{n \mathop = 1}^{k - 1} \left({1 - \frac {x^2} {n^2 \pi^2} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle - \frac {x^3} {k^2 \pi ^2} + O \left({x^5}\right)\) Big-O Notation

By Telescoping Series we find that the coefficient of $x^3$ in $P_k$ is given by:

$(1): \quad \displaystyle - \frac 1 {\pi^2} \sum_{i \mathop = 1}^k \frac 1 {i^2}$



From Euler Formula for Sine Function:

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

So by taking the limit as $k \to \infty$ in $(1)$ and equating with the coefficient of $x^3$ in the Power Series Expansion for Sine Function, we can deduce:

$\displaystyle - \frac 1 {\pi^2} \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = - \frac 1 {3!}$

hence:

$\displaystyle \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = \frac {\pi^2} 6$

$\blacksquare$


Proof 3

Let $x \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$ and let $n$ be a non-negative integer.

\(\displaystyle \frac {\cos \left({2n + 1}\right) x + i \sin \left({2n + 1}\right) x} {\sin^{2n + 1} x}\) \(=\) \(\displaystyle \frac {\left({\cos x + i \sin x}\right)^{2n + 1} } {\sin^{2n + 1} x}\) De Moivre's Formula
\(\displaystyle \) \(=\) \(\displaystyle \left({\cot x + i}\right)^{2n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 0}^{2n + 1} \binom{2n + 1} r i^r \cot^{2n + 1 - r} x\) Binomial Theorem
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\sin \left({2n + 1}\right) x} {\sin^{2n + 1} x}\) \(=\) \(\displaystyle \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} \left({-1}\right)^r \cot^{2 \left({n - r}\right)} x\) equating imaginary parts

Let $x_k = \dfrac {k \pi} {2n + 1}$ for $k = 1, 2, \ldots, n$.

Then:

$\sin \left({2n + 1}\right) x_k = 0$

So we have:

$\displaystyle \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} \left({-1}\right)^r \cot^{2 \left({n - r}\right)} x_k = 0$

for $k = 1, 2, \ldots, n$.

The numbers $x_k$ are all distinct and in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

By Shape of Cotangent Function, $\cot x$ is positive and injective in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Therefore $\cot^2 x$ is also injective in this interval.

Hence the numbers $c_k = \cot^2 x_k$ are distinct for $k = 1, 2, \ldots, n$.

These numbers are the $n$ distinct roots of the $n$th degree polynomial:

$\displaystyle f \left({c}\right) := \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} (-1)^r c^{n - r}$

By Viète's Formulas, we can calculate the sum of the roots:

$\displaystyle \sum_{k \mathop = 1}^n \cot^2 x_k = \frac {\binom {2n + 1} 3} {\binom {2n + 1} 1} = \frac {2 n \left({2 n - 1}\right)} 6$

From the Difference of Squares of Cosecant and Cotangent $\cot^2 x = \csc^2 x - 1$ we can similarly deduce:

$\displaystyle \sum_{k \mathop = 1}^n \csc^2 x_k = \frac {2 n \left({2 n - 1}\right)} 6 + n = \frac {2 n \left({2 n + 2}\right)} 6$

By Shape of Sine Function, $\sin x$ is positive in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

So from the Tangent Inequality $\sin x < x < \tan x$ for $x \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$, we can deduce that $\cot^2 x < 1 / x^2 < \csc^2 x$ in the same interval.

Summing this inequality from $x_1$ to $x_n$ gives:

\(\displaystyle \frac {2 n \left({2 n - 1}\right)} 6\) \(<\) \(\displaystyle \sum_{k \mathop = 1}^n \left({\frac {2 n + 1} {k \pi} }\right)^2\)
\(\displaystyle \) \(<\) \(\displaystyle \frac {2 n \left({2 n + 2}\right)} 6\)


or equivalently:

\(\displaystyle \frac {\pi^2} 6 \frac {2n \left({2n - 1}\right)} {\left({2n + 1}\right)^2}\) \(<\) \(\displaystyle \sum_{k \mathop = 1}^n \frac 1 {k^2}\)
\(\displaystyle \) \(<\) \(\displaystyle \frac {\pi^2} 6 \frac {2n \left({2n + 2}\right)} {\left({2n + 1}\right)^2}\)


By Combination Theorem for Limits of Functions, the left and right hand sides tend to $\dfrac {\pi^2} 6$ as $n$ tends to infinity.

Therefore by the Squeeze Theorem:

$\displaystyle \zeta \left({2}\right) = \sum_{k \mathop = 1}^\infty \frac 1 {k^2} = \frac {\pi^2} 6$

$\blacksquare$


Proof 4

From Sum of Reciprocals of Squares of Odd Integers,

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

Note that:

\(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^2} + \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + \frac {\pi^2} 8\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\displaystyle \frac {\pi^2} 6\)

$\blacksquare$


Proof 5

Let the function $f_n$ be defined as:

$(1): \quad f_n \left({x}\right) := \dfrac 1 2 + \cos x + \cos 2 x + \cdots + \cos n x$

By Sum of Cosines of Multiples of Angle:

$(2): \quad f_n \left({x}\right) = \dfrac {\sin \left({\left({2 n + 1}\right) x / 2}\right)} {2 \sin \left({x / 2}\right)}$

Let $E_n$ be defined as:

\(\displaystyle E_n\) \(=\) \(\displaystyle \int_0^\pi x f_n \left({x}\right) \mathrm d x\)
\((3):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 4 + \sum_{k \mathop = 1}^n \left({\frac {\left({-1}\right)^k - 1} {k^2} }\right)\) Primitive of $x \cos a x$ and algebra

The terms for even $k$ on the {right hand side are zero, so:

$\displaystyle (4): \quad \dfrac 1 2 E_{2 n - 1} = \dfrac {\pi^2} 8 - \sum_{k \mathop = 1}^n \dfrac 1 {\left({2 k - 1}\right)^2}$

It remains to be shown that:

$\displaystyle \lim_{n \mathop \to \infty} E_{2 n - 1} = 0$

which will establish:

$\displaystyle (5): \quad \dfrac {\pi^2} 8 = \sum_{k \mathop = 1}^n \dfrac 1 {\left({2 k - 1}\right)^2}$


Using $(2)$, let $g \left({x}\right)$ be the function defined as:

$g \left({x}\right) := \dfrac {\mathrm d} {\mathrm d x} \left({\dfrac {x / 2} {\sin \left({x /2}\right)} }\right)$

Using Integration by Parts, we obtain:

$\displaystyle (6): \quad E_{2 n - 1} = \dfrac 1 {4 n - 1} \left({2 + 2 \int_0^\pi g \left({x}\right) \cos \dfrac {\left({4 n - 1}\right) x} 2 \, \mathrm d x}\right)$

during which we use Limit of $\dfrac {\sin x} x$: Corollary:

$\displaystyle \lim_{x \mathop \to 0} \dfrac {x / 2} {\sin \left({x / 2}\right)} = 1$

We have that $g \left({x}\right)$ is increasing on the interval of integration.

Therefore $g \left({x}\right)$ is bounded on the interval $\left[{ 0 \,.\,.\, \pi}\right]$ by $g \left({\pi}\right) = \dfrac \pi 2$.

Hence $(5)$ has been established as being true.


Now we divide the (strictly) positive integers into even and odd, and use $(5)$ to obtain:

\(\displaystyle \sum_{k \mathop = 1}^\infty \frac 1 {k^2}\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty \frac 1 {\left({2 k}\right)^2} + \sum_{k \mathop = 1}^\infty \frac 1 {\left({2 k - 1}\right)^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 4 \sum_{k \mathop = 1}^\infty \frac 1 {k^2} + \frac {\pi^2} 8\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 3 4 \sum_{k \mathop = 1}^\infty \frac 1 {k^2}\) \(=\) \(\displaystyle \frac {\pi^2} 8\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{k \mathop = 1}^\infty \frac 1 {k^2}\) \(=\) \(\displaystyle \frac 4 3 \frac {\pi^2} 8\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 6\)

$\blacksquare$


Proof 6

\(\displaystyle \zeta \left({2}\right)\) \(=\) \(\displaystyle \left({-1}\right)^2 \dfrac {B_2 2^1 \pi^2} {2!}\) Riemann Zeta Function at Even Integers
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^2 \left({\dfrac 1 6}\right) \dfrac {2^1 \pi^2} {2!}\) Definition of Sequence of Bernoulli Numbers
\(\displaystyle \) \(=\) \(\displaystyle \left({\dfrac 1 6}\right) \left({\dfrac 2 2}\right) \pi^2\) Definition of Factorial
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^2} 6\) simplifying

$\blacksquare$


Proof 7

By Fourier Series of $x^2$, for $x \in \left[{-\pi \,.\,.\, \pi}\right]$:

$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \frac 4 {n^2} \cos n x}\right)$

Setting $x = \pi$:

\(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac{\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \frac 4 {n^2} \cos \pi x}\right)\)
\(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \left({-1}\right)^n \frac 4 {n^2} }\right)\) Cosine of Multiple of $\pi$
\(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac {\pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {2 \pi^2} 3\) \(=\) \(\displaystyle 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\displaystyle \frac {\pi^2} 6\)

$\blacksquare$


Proof 8

By Fourier Series of $x$, for $x \in \left({- \pi \,.\,.\, \pi}\right)$:

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


Hence:

\(\displaystyle \frac 1 \pi \int_{-\pi}^\pi x^2 \mathrm d x\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \left({\frac{2 \left({-1}\right)^{n + 1} } n}\right)^2\) Parseval's Theorem
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 2 \pi \int_0^\pi x^2 \mathrm d x\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 4 {n^2}\) Definite Integral of Even Function
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {2 \pi^2} 3\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 4 {n^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) \(=\) \(\displaystyle \frac {\pi^2} 6\)

$\blacksquare$


Proof 9

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

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

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

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


Setting $x = 0$:

\(\displaystyle f \left({0}\right)\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos 0} {n^2} + \left({\frac {\left({-1}\right)^n \pi} n + \frac {2 \left({\left({-1}\right)^n - 1}\right)} {\pi n^3} }\right) \sin 0}\right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({0 - \pi}\right)^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos 0} {n^2} }\right)\) Sine of Zero is Zero
\(\displaystyle \leadsto \ \ \) \(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) Cosine of Zero is One
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 2 - \frac {\pi^2} 3\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 6\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)

$\blacksquare$


Proof 10

From Mittag-Leffler Expansion for Hyperbolic Cotangent Function, we have:

$\displaystyle \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$

We can write:

\(\displaystyle \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z}\) \(=\) \(\displaystyle \frac 1 {2 z} \paren {\frac {\pi \paren {e^{\pi z} + e^{-\pi z} } } {e^{\pi z} - e^{-\pi z} } - \frac 1 z}\) Definition of Hyperbolic Cotangent
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 z} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {z \paren {e^{2 \pi z} - 1} } }\)

We therefore have:

$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }$

We have at $z = 0$:

\(\displaystyle \pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1\) \(=\) \(\displaystyle \pi \times 0 \times \paren {e^0 + 1} - e^0 + 1\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - e^0\)
\(\displaystyle \) \(=\) \(\displaystyle 0\) Exponential of Zero

and:

$2 z^2 \paren {e^{2 \pi z} + 1} = 0$

So by L'Hopital's Rule:

\(\displaystyle \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }\) \(=\) \(\displaystyle \lim_{z \mathop \to 0} \paren {\frac {\pi \paren {e^{2 \pi z} + 1} + 2 \pi^2 z e^{2 \pi z} - 2 \pi e^{2 \pi z} } {4 z \paren {e^{2 \pi z} - 1} + 4 \pi z^2 e^{2 \pi z} } }\) Product Rule, Derivative of Exponential Function, Derivative of Power
\(\displaystyle \) \(=\) \(\displaystyle \lim_{z \mathop \to 0} \paren {\frac {2 \pi^2 e^{2 \pi z} + 2 \pi^2 e^{2 \pi z} + 4 \pi^3 z e^{2 \pi z} - 4 \pi^2 e^{2 \pi z} } {4 \paren {e^{2 \pi z} - 1} + 8 \pi z e^{2 \pi z} + 8 \pi z e^{2 \pi z} + 8 \pi^2 z^2 e^{2 \pi z} } }\) L'Hopital's Rule
\(\displaystyle \) \(=\) \(\displaystyle 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {z e^{2 \pi z} } {-4 + e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} } }\)
\(\displaystyle \) \(=\) \(\displaystyle 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {e^{2 \pi z} + 2 \pi z e^{2 \pi z} } {2 \pi e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} + e^{2 \pi z} \paren {16 \pi z + 16 \pi} } }\) L'Hopital's Rule
\(\displaystyle \) \(=\) \(\displaystyle 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {1 + 2 \pi z} {16 \pi^3 z^2 + 32 \pi^2 z + 8 \pi + 16 \pi z + 16 \pi} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {4 \pi^3} {24 \pi}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 6\)

giving:

$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$

$\blacksquare$


Historical Note

The Basel Problem was first posed by Pietro Mengoli in $1644$.

Its solution is generally attributed to Leonhard Euler , who solved it in $1734$ and delivered a proof in $1735$.

However, it has also been suggested that it was in fact first solved by Nicolaus I Bernoulli.

Jacob Bernoulli had earlier established that the series was convergent, but had failed to work out what to.

The problem is named after Basel, the home town of Euler as well as of the Bernoulli family.


If only my brother were alive now.
-- Johann Bernoulli


Sources