I've always wanted to write an article and talk to you: $$\frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\ =\ \frac{π^2 }{6}$$ This article assumes that the reader loves mathematics and has mastered the high mathematics knowledge.

First we need to review the trigonometric functions.

For any angle $x $, we have ${\sin^2 x}+\cos^2x=1$, which is one thing with the Pythagorean theorem.

Next is an important formula, a little bit difficult, and drawing a picture might be understandable.

$$\sin (x+y) = \sin x \cos y + \cos x \sin y$$ Actually, we don't need to use this formula, we just need a special case:

When $x =y$, that is $$\sin (2x) = 2 \sin x \cos x$$ I believe that as long as it is a hobby of mathematics readers, look at these two pictures, will be able to prove this formula

! [Write a description of the picture here] (__http://img.blog.csdn.net/20160618170434781__)

Proof of $\sin (2x) = 2 \sin x \cos x$

! [Write a description of the picture here] (__http://img.blog.csdn.net/20160618170444359__)

Proof of $\sin (x+y) = \sin x \cos y + \cos x \sin y$

By drawing a trigonometric image,

We can also easily verify

$$\cos x = \sin (x+\frac{π}{2}) $$ and $$\sin (π-x) = \sin x$$

And then we can start proving that

(This certificate was taken from the American Journal of Mathematics, 109th February 2002 pp. 196-200 author department Josef Hofbauer)

Because $\sin (2x) = 2 \sin x \cos x$

So $\sin x= 2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) $

Take the reciprocal, square, get $$\frac{1}{\sin^2 x} = \frac{1}{4}\frac{1}{\sin^2 (X/2) \cos^2 (X/2)}$$ and then according to ${\sin^2 x}+\cos^2x=1$

We have

$$\frac{1}{\sin^2 x} = \frac{1}{4}\frac{\sin^2 (X/2) +\cos^2 (X/2)}{\sin^2 (X/2) \cos^2 (X/2)}=\frac{1}{4} (\frac{1}{\ Cos^2 (X/2)} + \frac{1}{\sin^2 (X/2)}) $$ next use nature, $\cos (X/2) = \sin ((x+π)/2) $

We get the core relational formula:

$$\frac{1}{\sin^2 x} = \frac{1}{4} (\frac{1}{\sin^2\frac{x}{2}} + \frac{1}{\sin^2\frac{x+π}{2}}) \ \ \ \ \ (*) $$ This is one of the most important steps in proving that we call this relationship "(*)"

Now we know by definition $\sin (Π/2) =\, \sin90°=1$

Then squared, take the reciprocal, and reuse (*) The formula, we have

\begin{split}

1 & = \frac{1}{\sin^2 (Π/2)} \ \

& =\frac{1}{4} (\frac{1}{\sin^2 (Π/4)} + \frac{1}{\sin^2 (3Π/4)}) \ \

& =\frac{1}{4^2} (\frac{1}{\sin^2 (Π/8)} + \frac{1}{\sin^2 (3Π/8)}+\frac{1}{\sin^2 (5Π/8)} + \frac{1}{\sin^2 (7Π/8)} )\\

& = \dots

\end{split} can do this all the time.

Now, using Identity $\sin (π-x) =\sin x$

You can get

\begin{split}

1 & =\frac{2}{4^2} (\frac{1}{\sin^2 (Π/8)} + \frac{1}{\sin^2 (3Π/8)}) \ \

& =\frac{2}{4^3} (\frac{1}{\sin^2 (Π/16)} + \frac{1}{\sin^2 (3Π/16)}+\frac{1}{\sin^2 (5Π/16)} + \frac{1}{\sin^2 (7π/ 16)}) \ \

& ={\frac{2}{4^4} (\frac{1}{\sin^2 (Π/32)} + \frac{1}{\sin^2 (3Π/32)}+\frac{1}{\sin^2 (5Π/32)} +\dots + \frac{1}{\ Sin^2 (15Π/32)})}\\

& = \dots

\end{split} We call this relationship "(* *)" type

Some readers may ask, why do this, in fact, the reason is immediately clear, the purpose is only one, so that all $\sin () $ in the value is a sharp angle.

Because for acute angle x we have $\sin x < x < \tan x$, see

! [Write a description of the picture here] (__http://img.blog.csdn.net/20160618170529602__)

Take the countdown, square, get

$$\frac{1}{\sin^2 x} > \frac{1}{x^2} > \frac{1}{\tan^2 x}$$ and we know

$$\frac{1}{\tan^2 x}=\frac{\cos^2 x}{\sin^2 x}=\frac{1-\sin^2 x}{\sin^2 x}=\frac{1}{\sin^2 x}-1$$ so

$$\frac{1}{\sin^2 x}-1<\frac{1}{x^2}<\frac{1}{\sin^2 x}$$

Now combine the previously deduced relationship (* *):

$$

1 ={\frac{2}{4^4} (\frac{1}{\sin^2 (Π/32)} + \frac{1}{\sin^2 (3Π/32)}+\frac{1}{\sin^2 (5Π/32)} +\dots + \frac{1}{\sin^2 ( 15Π/32)})}

$$

We can get the following unequal relationship

\begin{split}

1-\frac{2}{4^4}*2^3 & <{\frac{2}{4^4} (\frac{1}{(Π/32) ^2} + \frac{1}{(3Π/32) ^2}+\frac{1}{(5Π/32) ^2} +\dots + \ frac{1}{(15Π/32) ^2})} < 1\\

1-\frac{2}{4^4}*2^3 & < {\frac{2}{4^4}* 4^5 (\frac{1}{π^2} + \frac{1}{(3π) ^2}+\frac{1}{(5π) ^2} +\dots + \frac{1}{ (15π) ^2})} <1\\

1-\frac{1}{2^4}& < {8 (\frac{1}{π^2} + \frac{1}{(3π) ^2}+\frac{1}{(5π) ^2} +\dots + \frac{1}{(15π) ^2})} <1

\end{split}

(Please note that these three unequal relationships may take time to read carefully, which is the most difficult step in the full text, and hopefully the reader will be able to understand the logic of each line patiently.) ）

By observing, we can find that, in the previous (* *), we only use $$1 = {\frac{2}{4^4} (\frac{1}{\sin^2 (Π/32)} + \frac{1}{\sin^2 (3Π/32)}+\frac{1}{\sin^2 (5π /32)} +\dots + \frac{1}{\sin^2 (15Π/32)})}$$

If in the first step, use more (*) several times,

Causes the number of items in the $\sin$ (* *) to grow from $8=2^3$ to $2^n$

Then there are

$$1-\frac{1}{2^{n+1}} < {8 (\frac{1}{π^2} + \frac{1}{(3π) ^2}+\frac{1}{(5π) ^2} +\dots +

\frac{1}{((2^{n+1}-1) π) ^2})}<1$$

So, when n is large, \frac{1}{2^{n+1}} can be ignored and we have

$${\,8\, (\frac{1}{π^2} + \frac{1}{(3π) ^2}+\frac{1}{(5π) ^2}+\frac{1}{(7π) ^2} +\dots +

)}= 1$$

i.e. $$\frac{1}{1^2}+\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +\cdots =\frac{π^2}{8} $$

Now we're only one step away from the conclusion.

Make $\zeta (2) = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + \dots$

So $\zeta (2)/4 = 1/2^2 + 1/4^2 + 1/6^2 + 1/8^2 + 1/10^2 + \dots$

Subtract two, you can get

$\zeta (2)-\zeta (2)/4= 1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + \dots=π^2/8$

So $3\zeta (2)/4 =π^2/8$, Seek $\zeta (2) =π^2/6$, that is, we want to testify the conclusion:

$$\FRAC{1}{1^2}+\FRAC{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\ =\ \frac{π^2}{6}$$

How about, fun, math is always like this, with the most ingenious logic chain to construct the most beautiful proof.

As long as there is a little curiosity, and enough patience, everyone can enjoy the fun of maths.

I wish you a pleasant summer vacation.

Jabo Name

June 20, 2014 at Columbus, Ohio, USA

(Last updated on June 18, 2016, thank Sun Ho for reading the first draft of this article.)

Zeta (2)