\documentclass[a4paper]{article}
\usepackage{yhmath}
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\usepackage{xcolor}
\usepackage{amssymb}
\usepackage{tikz}
\begin{document}
\title{Basel Problem Made Simple}
\author{Amit Yaron}
\maketitle
\section*{Introduction}
What does "made simple" mean? Maybe, that you need less knowledge in order to
understand the solution. For example, a solution to the Basel problem, that is
the sum
\[
\sum_{n=0}^{\infty}\frac1{n^2}
\]
using the Fourier series as follows:
\begin{align*}
 x^2=\frac1{2\pi}\int_{-\pi}^{\pi}x^2+\frac1{\pi}\sum_{n=1}^{\infty}(\cos(nx)\int_{-\pi}^{\pi}x^2\cos(nx)dx+sin(nx)\int_{-\pi}^{\pi}x^2\sin(nx)dx)
\end{align*}
Since $\sin$ is an odd function:
\[
 \forall n\in\mathbb{Z},\quad \int_{-\pi}^{\pi}x^2\sin(nx)dx=0
\]
and
\[
 \frac1{2\pi}\int_{-\pi}^{\pi}x^2dx=\frac{x^3}{6\pi}\bigg|_{-\pi}^{\pi}=\frac{\pi^2}3
\]
and
\begin{align*}
 \frac1{\pi}\int_{-\pi}^{\pi} x^2\cos(nx) dx&=\frac{x^2\sin(nx)}{n\pi}
      \bigg|_{-pi}^{\pi}-\frac{1}{n\pi}\int_{-\pi}^{\pi}2x\sin(nx)dx\\
 &=-\frac{1}{n\pi}\int_{-\pi}^{\pi}2x\sin(nx)dx\\
 &=\frac{2x\cos(nx)}{n^2\pi}\bigg|_{-pi}^{\pi}+
      \frac{2}{n^2\pi}\int_{-\pi}^{\pi}\cos(nx)dx\\
 &=\frac{2\pi\cos(n\pi)-2(-\pi)\cos(-n\pi)}{n^2\pi}{n^2\pi}+
   \frac{\sin(nx)}{n^3\pi}\bigg|_{-\pi}^{\pi}\\
 &=\frac{4\cos(n\pi)}{n^2}\\
 &=\frac{4(-1)^n}{n^2}
\end{align*}
Thus, we get that in $(-\pi,\pi)$:
\[
 x^2=\frac{\pi^2}3+\sum_{n=1}^{\infty}\frac{4(-1)^n(cos(nx)}{n^2}
\]
Plug $x=\pi$, and get:
\begin{align*}
 &\pi^2=\frac{\pi^2}3+\sum_{n=1}^{\infty}\frac{4(-1)^n\cos(n\pi)}{n^2}
       =\frac{\pi^2}3+\sum_{n=1}^{\infty}\frac{4}{n^2}\Rightarrow\\
 \Rightarrow& \sum_{n=1}^{\infty}\frac{4}{n^2}=\frac{2\pi^2}3\Rightarrow\\
 \Rightarrow& \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}6
\end{align*}

But how much math should you learn to use them Fourier series? So,
in this article I'm going to show a method using
Euclidean geometry.
\section{Behind the Solution - Attenuation of Light}
The brightness of light that comes from a light source is proportional to
the inverse square of the distance from it. Suppose that along an infinite
line, lightsources are placed every two units of distance, what is the
brightness in the middle between two light sources? The distances between
a middle point to the rest of the light sources are: 1,3,5,... and there
are two light sources for each distance, so the brightness is proportional
to
\begin{equation*}
 \sum_{n=-\infty}^{\infty}\frac1{(2n+1)^2}=2\sum_{n=0}^{\infty}\frac1{(2n+1)^2}
\end{equation*}
To find the sum, one can view the infinite straight line as a limit of a
sequence of circles, or half-circles, or quarter-circles and so on.
\section{A Sequence of Circles}
For each natural $n$, let us build a circle whose circumferencse is $2^{n+1}$,
and place $2^n$ equally spaced points along it. So, the are between 2
consecutive points is 2, and the radius of the circle is $\frac{2^n}{\pi}$. Following is an example with $n=3$:\\ \\
\setlength{\unitlength}{1mm}
\begin{picture}(40,40)
{\put(25,25){\circle{26}}}
\put(37.01,29.97){\line(0.92,0.38){0.92}}
\put(37.93,30.35){$P_1$}
\put(29.97,37.01){\line(0.38,0.92){0.38}}
\put(30.35,37.39){$P_2$}
\put(20.03,37.01){\line(-0.38,0.92){0.38}}
\put(16.05,37.89){$P_3$}
\put(12.99,29.97){\line(-0.92,0.38){0.92}}
\put(8.57,30.59){$P_4$}
\put(12.99,20.03){\line(-0.92,-0.38){0.92}}
\put(8.57,19.11){$P_5$}
\put(20.03,12.99){\line(-0.38,-0.92){0.38}}
\put(16.05,10.07){$P_6$}
\put(29.97,12.99){\line(0.38,-0.92){0.38}}
\put(30.35,10.07){$P_7$}
\put(37.01,20.03){\line(0.92,-0.38){0.92}}
\put(37.93,19.11){$P_8$}
\end{picture}
\\Now, let us add a point on the in the middle between two consecutive points,
call it Q, and draw chords from it to all the other points.   \\
\begin{picture}(40,40)
\put(25,25){\circle{26}}
\put(37.01,29.97){\line(0.92,0.38){0.92}}
\put(37.93,30.35){$P_1$}
\put(29.97,37.01){\line(0.38,0.92){0.38}}
\put(30.35,37.39){$P_2$}
\put(20.03,37.01){\line(-0.38,0.92){0.38}}
\put(16.05,37.89){$P_3$}
\put(12.99,29.97){\line(-0.92,0.38){0.92}}
\put(8.57,30.59){$P_4$}
\put(12.99,20.03){\line(-0.92,-0.38){0.92}}
\put(8.57,19.11){$P_5$}
\put(20.03,12.99){\line(-0.38,-0.92){0.38}}
\put(16.05,10.07){$P_6$}
\put(29.97,12.99){\line(0.38,-0.92){0.38}}
\put(30.35,10.07){$P_7$}
\put(37.01,20.03){\line(0.92,-0.38){0.92}}
\put(37.93,19.11){$P_8$}
\put(38,25){\line(1,0){1}}
\put(39,24){$Q$}
\put(38,25){\line(-0.99,4.97){0.99}}
\put(38,25){\line(-8.03,12.01){8.03}}
\put(38,25){\line(-18.22,12.01){18.22}}
\put(38,25){\line(-25.01,4.97){25}}
\put(38,25){\line(-0.99,-4.97){0.99}}
\put(38,25){\line(-8.03,-12.01){8.03}}
\put(38,25){\line(-18.22,-12.01){18.22}}
\put(38,25){\line(-25.01,-4.97){25}}
\end{picture}
\\
Let use write the length of the counterclockwise arc from A to B as
$\wideparen{AB}$
So, for $k=1,2,\ldots,2^n$:
\[
 \min\{\wideparen{QP_k},\wideparen{P_kQ}\}=2k-1
\]
But, I guess you can barely see that the chord $QP_1$ has been added to the circle.
The lengths of the cords gett closer to those of the arcs.
Let us sum the inverse squares of the chords and use the squeeze theorem.
\section{Sum of Inverse Squares of Chords}
\[
 \overline{QP_k}<\min\{\wideparen{QP_k},\wideparen{P_kQ}\}=2k-1\Rightarrow
 \overline{QP_k}^2<(2k-1)^2\Rightarrow\overline{QP_k}^{-2}>(2k-1)^{-2}
\]
Let us start calculating the sum of inverse square with $n=2$, the radius
of the circle will be $\frac2{\pi}$\\\\
\begin{picture}(10,10)
\put(5,5){\circle{7}}
\put(5,8.5){\line(0,1){1}}
\put(5,1.5){\line(0,-1){1}}
\put(8.5,5){\line(1,0){1}}
\put(4,9.5){$P_1$}
\put(4,-1.5){$P_2$}
\put(9.5,4.5){$Q$}
\put(5,8.5){\line(0,-7){7}}
\put(8.5,5){\line(-1,1){3.5}}
\put(8.5,5){\line(-1,-1){3.5}}
\end{picture}
\\\\
$\angle{P_1QP_2}$ is an inscribed angle opposite the diameter, and
$\overline{QP_1}=\overline{QP_2}$\\
and $\overline{P_1P_2}=\frac{4}{\pi}$
\\ Thus, from the Pythagorean theorem:
\begin{align*}
 &\overline{QP_1}^2+\overline{QP_2}^2=\frac{16}{\pi^2}\Rightarrow\\
 \Rightarrow &2\overline{QP_1}^2=\frac{16}{\pi^2}\Rightarrow\\
 \Rightarrow &\overline{QP_1}^2=\overline{QP_2}^2=\frac{8}{\pi^2}\Rightarrow\\
 \Rightarrow &\overline{QP_1}^{-2}=\overline{QP_2}^{-2}=\frac{\pi^2}8\Rightarrow\\
 \Rightarrow &\overline{QP_1}^{-2}+\overline{QP_2}^{-2}=\frac{\pi^2}4
\end{align*}
For values of $n$ greater than 2, let us sum up the inverse square of
chords in pairs: be $k$ a natural number, such that $1\le k\le2^{n-1}$.
Then if we take the chords $QP_k$ and $QP_{2^{n-1}+k}$ and connect $P_k$
and $P_{2^{n-1}+k}$. The length of arc $\wideparen{P_kP_{2^{n-1}+k}}$ is
half the circumfernce of the circle. Thus, $\angle{P_kQP_{\cdots}}=\frac{\pi}2$.
\\ \\
\begin{picture}(40,40)
\put(25,25){\circle{26}}
\put(37.01,29.97){\line(0.92,0.38){0.92}}
\put(37.93,30.35){$P_1$}
\put(29.97,37.01){\line(0.38,0.92){0.38}}
\put(30.35,37.39){$P_2$}
\put(20.03,37.01){\line(-0.38,0.92){0.38}}
\put(16.05,37.89){$P_3$}
\put(12.99,29.97){\line(-0.92,0.38){0.92}}
\put(8.57,30.59){$P_4$}
\put(12.99,20.03){\line(-0.92,-0.38){0.92}}
\put(8.57,19.11){$P_5$}
\put(20.03,12.99){\line(-0.38,-0.92){0.38}}
\put(16.05,10.07){$P_6$}
\put(29.97,12.99){\line(0.38,-0.92){0.38}}
\put(30.35,10.07){$P_7$}
\put(37.01,20.03){\line(0.92,-0.38){0.92}}
\put(37.93,19.11){$P_8$}
\put(38,25){\line(1,0){1}}
\put(39,24){$Q$}
\put(38,25){\line(-8.03,12.01){8.03}}
\put(38,25){\line(-18.22,-12.01){18.22}}
\put(29.87,37.01){\line(-10.19,-24.02){10.19}}
\end{picture}
\\
Now, let us add a figure of a right-angled triangle with its hypothenuse as
its base, draw a height, and talk about the inverse Pythagorean theorem.
\\ \\
\begin{picture}(40,25)
\put(5,5){\line(1,0){30}}
\put(35,5){\line(-19.2,14.4){19.2}}
\put(15.8,19.4){\line(-10.8,-14.4){10.8}}
\put(15.8,19.4){\line(0,-14.4){14.4}}
\put(17.72,17.96){\line(-1.08,-1.44){1.2}}
\put(16.64,16.52){\line(-1.92,1.44){2}}
\put(15.8,7){\line(1,0){2}}
\put(17.8,7){\line(0,-1){2}}
\put(19,3.3){$c$}
\put(8.9,12.2){$a$}
\put(25,12.2){$b$}
\put(15.9,11.2){$h$}
\end{picture}
\\
Now, according to the Pythagorean theorem:
\[
 a^2+b^2=c^2
\]
And from the formula for the area of a triangle:
\[
 ab=ch\Rightarrow a^2b^2=c^2h^2=(a^2+b^2)h^2
\]
Dividing by $a^2b^2h^2$, we get that:
\[
 h^{-2}=a^{-2}+b^{-2}
\]
That's the inverse Pythagorean theorem.\\
Let us now draw a circle with the triangle inscribed in it, and add some points and a radius.\\
\begin{picture}(40,35)(0,-10)
\put(5,5){\line(1,0){30}}
\put(35,4){$P_k$}
\put(35,5){\line(-19.2,14.4){19.2}}
\put(15,20){$Q$}
\put(15.8,19.4){\line(-10.8,-14.4){10.8}}
\put(1,4){$P_{\cdots}$}
\put(15.8,19.4){\line(0,-14.4){14.4}}
\put(17.72,17.96){\line(-1.08,-1.44){1.2}}
\put(16.64,16.52){\line(-1.92,1.44){2}}
\put(15.8,7){\line(1,0){2}}
\put(17.8,7){\line(0,-1){2}}
\put(20,5){\circle{30}}
\put(12.9,11.2){$h_k$}
\put(20,5){\line(-4.2,14.4){4.2}}
\put(19,2){$O$}
\end{picture}
\\
So, the sum of inverse squares of chords become a sum of inverse squares of
heights.  \\
Let $h_k$ be the height of $\triangle P_{2^{n-1}+k}QP_k$ in circle $n$, then $s_n$ the sum of inverse squares of chords in circle $n$ will be:
\[
 s_n=\sum_{k=1}^{2^n}(\overline{QP_k})^{-2}=
\sum_{k=1}^{2^{n-1}}(\overline{QP_k})^{-2}+
(\overline{QP_{2^{n-1}+k}})^{-2}=\sum_{k=1}^{2^{n-1}}h_k^{-2}
\]
$O$ is the center of the circle, and from the properties of inscribed angles,
we get that
\[
 \angle QP_{\cdots}P_k=\frac12\angle QOP_k
\]
and
\[
 \angle QP_kP_{\cdots}=\frac12\angle QOP_{\cdots}
\]
Those inscribed angles will be the central angles in the next circle, which
will have twice the number of points, a diameter twice as long as that of the
current circle, and the angles will be half of those of the current circle.
\\ Let $r_n$ be the radius of circle $n$.\\
Now, I would like to draw circle $n+1$, place circle $n$ inside it, and use
the proportionality theorem to find $s_{n+1}$ based on $s_n$.
In the bigger circle the poits will be:
\[
Q',P'_1,P'_2,\ldots,P'_{2^{n+1}}
\]
and the heights will be
\[
h'_1,h'_2,\ldots,h'_{2^n}
\]
\\
\begin{picture}(80,50)(0,-15)
 \put(40,10){\circle{60}}
% Left circle
\put(55,10){\circle{30}}
\put(40,10){\line(1,0){30}}
\put(70,9){$P_k$}
\put(70,10){\line(-19.2,14.4){19.2}}
\put(50,25){$Q$}
%\put(50.8,24.4){\line(-10.8,-14.4){10.8}}
\put(58,34){\line(-10.8,-14.4){18}}
% Add bigger height
\put(58,34){\line(0,-1){24}}
% Add Q'
\put(57.5,33.7){$Q'$}
% Add h'_k
\put (56,22){$h'_k$}
% Add A and B
\put(49,7){$A$}
\put(57.5,7){$B$}
\put(36,9){$P_{\cdots}$}
\put(50.8,24.4){\line(0,-14.4){14.4}}
\put(52.72,22.96){\line(-1.08,-1.44){1.2}}
\put(51.64,21.52){\line(-1.92,1.44){2}}
\put(50.8,12){\line(1,0){2}}
\put(52.8,12){\line(0,-1){2}}
\put(47.9,16.2){$h_k$}
\put(55,10){\line(-4.2,14.4){4.2}}
\put(54,7){$O$}
\end{picture}
\\ \\
What have we here?\\
\begin{align*}
&\angle QP_{2^{n-1}+k}A=\angle Q'P_{2^{n-1}+k}B\\
&\angle QAP_{2^{n-1}+k}=\angle Q'BP_{2^{n-1}+k}
\end{align*}
Two similar triangles! Thus, by proportional sections:
\[
 \frac{\overline{QP_{2^{n-1}+k}}}{h_k}=\frac{r_{n+1}}{h'_k}=\frac{2r_n}{h'_k}
\]
Let us square both sides:
\begin{equation}
\frac{\overline{QP_{2^{n-1}+k}}^2}{{h_k}^2}=\frac{4{r_n}^2}{{h'_k}^2}
\end{equation}
Now, let us move the circle to the other side:
\\
\begin{picture}(80,50)(0,-5)
 \put(40,10){\circle{60}}
% Right circle
\put(25,10){\circle{30}}
\put(10,10){\line(1,0){30}}
\put(40,9){$P_k$}
\put(40,10){\line(-19.2,14.4){24}}
\put(20,25){$Q$}
\put(20.8,24.4){\line(-10.8,-14.4){10.8}}
%\put(28,34){\line(-10.8,-14.4){18}}
% Add bigger height
\put(16,28){\line(0,-1){18}}
% Add Q'
\put(14.2,28.3){$Q'$}
% Add h'_k
\put (14.2,17.2){$h'_{\cdots}$}
% Add A and B
\put(19,7){$A$}
\put(14.2,7){$B$}
\put(6,9){$P_{\cdots}$}
\put(20.8,24.4){\line(0,-14.4){14.4}}
\put(22.72,22.96){\line(-1.08,-1.44){1.2}}
\put(21.64,21.52){\line(-1.92,1.44){2}}
\put(20.8,12){\line(1,0){2}}
\put(22.8,12){\line(0,-1){2}}
\put(17.9,16.2){$h_k$}
\put(25,10){\line(-4.2,14.4){4.2}}
\put(24,7){$O$}
\end{picture}
\\ \\
\\ \\ \\
Here, again we have two similar triangles:
\begin{align*}
&\angle QP_kA=\angle Q'P_kB\\
&\angle QAP_k=\angle Q'BP_k
\end{align*}
Thus, by proportional sections:
\[
 \frac{\overline{QP_{k}}}{h_k}=\frac{r_{n+1}}{h'_{2^{n-1}+k}}=\frac{2r_n}{h'_{2^{n-1}+k}}
\]
Let us square both sides:
\begin{equation}
\frac{\overline{QP_{k}}^2}{{h_k}^2}=\frac{4{r_n}^2}{({h'_{2^{n-1}+k})}^2}
\end{equation}
Let us sum (1) and (2):
\[
 \frac{4{r_n}^2}{{h'_k}^2}+\frac{4{r_n}^2}{({h'_{2^{n-1}+k})}^2}=
   \frac{\overline{QP_{2^{n-1}+k}}^2}{{h_k}^2}+\frac{\overline{QP_{k}}^2}{{h_k}^2}
\]
And because $\angle{P_kQP_{2^{n-1}}}$ is a right angle, and the radius of
circle $n$ is $r_n$:
\[
 \overline{QP_{2^{n-1}+k}}^2+\overline{QP_{k}}^2={(2r_n)}^2=4r_n^2
\]
Thus,
\begin{align*}
 &\sum_{k=1}^{2^n}\frac{4r_n^2}{h'_k}\\
 =&\sum_{k=1}^{2^{n-1}}\frac{4{r_n}^2}{{h'_k}^2}+\frac{4{r_n}^2}{({h'_{2^{n-1}+k})}^2}\\
 =&\sum_{k=1}^{2^{n-1}}\frac{4{r_n}^2}{h_k^2}
\end{align*}
and, by dividing the sides of the equation above by $4r_n^2$, we get that
the sum of inverse squares of chords in circle $n+1$ is:
\[
 s_{n+1}=\sum_{k=1}^{2^{n-1}}h_k^{-2}=s_n
\]
and
\[
 s_1=\frac{\pi^2}4
\]
Thus,
\[
 \lim_{n\to\infty}s_n=\frac{\pi^2}4
\]
But, the arcs of the sectors are longer than the heights. Thus the sum of
inverse squares of arcs is smaller than the sum of inverse squares oh
chords/heights. So, lines longer than the arcs are required to complete the
work.
\section{The Squeeze Theorem}
We need a squence $\{a_n\}$ of numbers such that:
\[
 \frac{\pi^2}4\geq\sum_{k=-2^{n-1}+1}^{2^{n-1}+1}\frac1{(2k+1)^2}\geq a_n
\]
and
\[
 \lim_{n\to\infty}a_n=\frac{\pi^2}4
\]
which will leave $\lim\limits_{n=-\infty}\limits^{\infty}\frac1{(2n+1)^2}$
no choice but to be $\frac{\pi^2}4$.
\\
Let us draw a sector inside a right-angles triangle whose base is of length
equal to the radius of the sector.
\\ \\
\begin{tikzpicture}
 \draw (1,0) -- (7,0);
 \draw (7,0) arc (0:20:60mm);
 \draw (6.638,2.052) -- (1,0);
 \node (O) at (0.75,0) {O};
 \draw (6.638,2.052) -- (6.638,0);
 \draw (7,0) --(7,2.163) -- (6.638,2.052);
 \node (A) at (6.63,-0.25) {A};
 \node (B) at (7,-0.25) {B};
 \node (C) at (7,2.27) {C};
 \node (D) at (6.63,2.17) {D};
\end{tikzpicture}
\\
"O" is the center of a circle.\\
Let:
\[
 r=\overline{OD}=\overline{OB}
\]
\[
 l=\wideparen{BD}
\]
and
\[
 h=\overline{DA}
\]
Then,
\[
 \overline{OA}=\sqrt{r^2-h^2}
\]
and using proportional sections:
\[
 \overline{BC}=\frac{rh}{\sqrt{r^2-h^2}}
\]
Using the areas of shapes that are one inside another:
\[
 \frac{rh}2<\frac{rl}2<\frac{r^2h}{2\sqrt{r^2-h^2}}
\]
Multiplying by $\frac2r$:
\[
 h<l<\frac{hr}{\sqrt{r^2-h^2}}\Rightarrow h^{-2}>l^{-2}>\frac{r^2-h^2}{h^2r^2}=
 h^{-2}-r^{-2}
\]
Now, let us have a look at circle $n+1$:\\
It's radius is $\frac{2^{n+1}}{\pi}$.\\
There are $2^n$ heights whose inverse squares are to be summed up. Let us call them
\[
 h_1,h_2,\ldots,h_2^n
\]
For convenience, let us define:
$$
l_k=\begin{cases}
 \wideparen{QP_k}=2k-1\quad&k\leq 2^{n-1}\\
 \wideparen{P_{2^n+k}Q}=2(2^n-k)+1\quad&otherwise
\end{cases}
$$
(If you want to understand it better, look at the example on page 3 with $m=2$)
So,
\[
 h_k^{-2}>l_k^{-2}>h_k^{-2}-\frac{\pi^2}{4^{n+1}}\quad k=1,2,\ldots,2^n
\]
Let us sum them up:
\[
 \sum_{k=1}^{2^n}h_k^{-2}>\sum_{k=1}^{2^n}l_k^{-2}>\sum_{k=1}^{2^n}h_k^{-2}-\frac{\pi^2}{4^{n+1}}
\]
Thus,
\[
 \frac{\pi^2}4>\sum_{k=1}^{2^n}\frac2{(2k-1)^2}>\frac{\pi^2}4-\frac{\pi^2}{4\cdot2^n}
\]
Taking limits:
\[
 \frac{\pi^2}4\geq\lim_{n\to\infty}\sum_{k=1}^{2^n}\frac2{(2k-1)^2}\geq
 \frac{\pi^2}4-\overbrace{\frac{\pi^2}{4\cdot2^n}}^{=0}=\frac{\pi^2}4
\]
Thus, per the squeeze theorem:
\[
 \lim_{n\to\infty}\sum_{k=1}^{2^n}\frac2{(2k-1)^2}=\frac{\pi^2}4\Rightarrow
 \lim_{n\to\infty}\sum_{k=1}^{2^n}\frac1{(2k-1)^2}=\frac{\pi^2}8
\]
And because
\[
\sum_{k=1}^n\frac1{(2k-1)^2}
\]
is increasing in $n$, we get that:
\[
 \sum_{n=1}^{\infty}\frac1{(2k-1)^2}=\frac{\pi^2}8
\]
\section{Adding the Inverse Squares of Even Number}
The last step in the calculation is to add the inverse squares of even numbers.
\[
 \sum_{k=1}^n \frac1{(2k)^2}
\]
is increasing, and is bounded because:
\[
 \sum_{k=1}^n \frac1{(2k)^2}<\sum_{k=1}^n \frac1{(2k-1)^2}
\]
Thus, $\sum\limits_{k=1}\limits^{\infty}\frac1{(2k)^2}$ converges.
Let
\[
 s=\sum_{n=1}^{\infty}\frac1{k^2}
\]
Then
\[
 \sum_{n=1}^{\infty}\frac1{(2k)^2}=\sum_{n=1}^{\infty}\frac1{4k^2}=\frac14\sum_{n=1}^{\infty}\frac1{k^2}=\frac{s}4
\]
Let us finish the calculation:
\begin{align*}
&s=\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n+1)^2}+\frac1{(2n)^2}=\frac{\pi^2}8+\frac{s}4\Rightarrow\\
\Rightarrow & 4s=\frac{\pi^2}2+s\Rightarrow\\
\Rightarrow & 3s=\frac{\pi^2}2\Rightarrow\\
\Rightarrow &s=\frac{\pi^2}6
\end{align*}
\section{Summary}
No advanced math courses are required to solve the Basel problem: Euclidean
geometry and basic knowledge of limits are enough. Here, I haven't even used
trigoneometric functions.\\
Go back to the basic to show some thinking out of the box.
\end{document}

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