Basel Problem

The Basel Problem asks for the precise sum of the infinite sum of the reciprocal of squares or: $\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + ... = ?$

The answer has been proven to be $\frac{\pi^2}{6}.$

Proof

The derivative of a function $f(x)$ is defined as $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$ So the derivative of $f(x) = e^x$ is $\lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}.$ Let $e^h - 1 = y.$ Then as $h \to 0$ , $y \to 0,$ and $h = \ln(1+y).$ $e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \lim_{y \to 0} \frac{y}{\ln(1+y)} = e^x \lim_{y \to 0} \frac{1}{\ln(1+y)^{1/y}} = e^x \cdot \frac{1}{\ln \left( \lim_{y \to 0} (1+y)^{1/y} \right)} = e^x \cdot \frac{1}{\ln e} = e^x.$ Since the derivative of $e^x$ is $e^x$ itself, the $n$ th derivative of $e^x$ will be $e^x.$ Now, consider the infinite sum $P(x) = \sum_{n=0}^{\infty} a_nx^n = a_0 + a_1x + a_2x^2 + \dots$ The derivatives at $x=0$ are $P(0) = a_0,$ $P'(0) = a_1,$ $P''(0) = 2a_2,$ $P'''(0) = 3 \cdot 2a_3,$ and so on, with $P^{(n)}(0) = n!a_n.$ Let this sum equal the function $f(x) = e^x.$ Then we must have $f^{(n)}(0) = P^{(n)}(0),$ which means $n!a_n = f^{(n)}(0) = e^0 = 1,$ so $a_n = \frac{1}{n!}.$ The series for $e^x$ is thus $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ Now consider the function $f(x) = \frac{\cos(x) + i\sin(x)}{e^{ix}}.$ We can evaluate its derivative:

$f'(x) = \frac{i(\cos(x) + i\sin(x))e^{ix} - (\cos(x) + i\sin(x))(ie^{ix})}{(e^{ix})^2} = 0$

This means $f(x)$ is a constant. Since $f(0) = 1,$ $f(x) = 1$ for all $x,$ or $e^{ix} = \cos(x) + i\sin(x).$ We have derived Euler's Formula.

Using the infinite series expansion and substituting $ix$ for $x,$ we get: $e^{ix} = \sum_{n=0}^{\infty} \frac{(ix)^n}{n!} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \dots$ $= 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} + \dots$ $= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \right) + i \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \right).$ By equating the real and imaginary parts of the series with Euler's formula, we find the series expansions for sine and cosine: $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$ $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$ Since $\sin(n\pi) = 0$ for all integers $n,$ the roots of the function $\sin(x)/x$ are at $x = \pm \pi, \pm 2\pi, \pm 3\pi, \dots$ We can factor the polynomial expansion in terms of its roots: $\sin(x) = x( 1 - \frac{x^2}{\pi^2})( 1 - \frac{x^2}{(2\pi)^2})( 1 - \frac{x^2}{(3\pi)^2}) \dots$ Now, consider the coefficient of the $x^3$ term in the expansion of $\sin(x).$ From the infinite product, the coefficient of the $x^3$ term is found by multiplying the $x$ with each term with the $-(x^2/n^2\pi^2):$ $\text{Coefficient of } x^3 = -\left( \frac{1}{\pi^2} + \frac{1}{(2\pi)^2} + \frac{1}{(3\pi)^2} + \dots \right) = -\frac{1}{3!}$ Equating the two expressions for the $x^3$ coefficient, we get: $-\frac{1}{6} = -\frac{1}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2}$ $\implies \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$

~grogg007

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Basel Problem

Proof