proof of the Basel problem

[autumn charm]

Let $n \geq 1$ be a natural number and $\theta \notin \pi \mathbb{Z}$ then using demoivre’s $sin((2n+1)\theta)=Im\left(((cos(\theta)+isin(\theta))^{2n+1}\right)=Im\left(\sum_{k=0}^{2n+1} \binom{2n+1}{k} i^{k} sin(\theta)^{k} cos(\theta)^{2n+1-k}\right)=\sum_{k=0}^{n} \binom{2n+1}{2k+1} (-1)^{k} sin(\theta)^{2k+1} cos(\theta)^{2n-2k}$ dividing by $sin(\theta)^{2n+1}$ on each side we get that $\frac{sin((2n+1)\theta)}{sin(\theta)^{2n+1}}=\sum_{k=0}^{n} \binom{2n+1}{2k+1} (-1)^{k} cotan(\theta)^{2(n-k)}$ now in that case if we consider the polynomial $P=\sum_{k=0}^{n} \binom{2n+1}{2k+1} (-1)^{k} X^{n-k}$ then the roots of $P$ are ${cotan(\frac{k\pi}{2n+1})^{2}, k \in {1,…,n}}$ in that case using the relation between the roots and the coefficients of $P$ we have that $\frac{1}{2n+1} \binom{2n+1}{3}=\sum_{k=1}^{n} cotan(\frac{k\pi}{2n+1})^{2}$ thus $\frac{n(2n-1)}{3}=\sum_{k=1}^{n} cotan(\frac{k\pi}{2n+1})^{2}$ now $\sum_{k=1}^{n} \frac{1}{sin^{2}(\frac{k\pi}{2n+1})}=\frac{n(2n-1)}{3}+n=\frac{2n(n+1)}{3}$ (because $1/sin^2(t)=1+cotan^2(t)$) now we can prove that for all $t>0$, $tan(t) \geq t \geq sin(t)$ (using the mean value inequality and the convexity of tan, for instance) so we have that $\forall k \in {1,…,n}, tan(\frac{k\pi}{2n+1})^{2} \geq \frac{k^{2} \pi^{2}}{(2n+1)^{2}} \geq sin(\frac{k \pi}{2n+1})^{2}$, so inverting on each side and summing we have that $\sum_{k=1}^{n} cotan(\frac{k \pi}{2n+1})^{2} \le \frac{(2n+1)^{2}}{\pi^{2}} \sum_{k=1}^{n} \frac{1}{k^{2}} \le \sum_{k=1}^{n} \frac{1}{sin(\frac{k \pi}{2n+1})^{2}}$ ie $\frac{n(2n-1)}{3} \le \frac{(2n+1)^{2}}{\pi^{2}} \sum_{k=1}^{n} \frac{1}{k^{2}} \le \frac{2n(n+1)}{3}$ so finally $\pi^{2} \frac{n(2n-1)}{3(2n+1)^{2}} \le \sum_{k=1}^{n} \frac{1}{k^{2}} \le \pi^{2} \frac{2n(n+1)}{3(2n+1)^{2}}$ both sides tend to $\frac{\pi^{2}}{6}$ when $n$ approaches infinity, so using the squeeze theorem we have that $\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}$

[supple lantern]

Rotor 😑

[autumn charm]

@boreal portal here’s a proof if you want

I really can’t format latex correctly I’m very sorry

[boreal portal]

Ah, no worries, thanks! I'll take a look at it later.
If you want, we can hop on in VC in the afternoon and format this in a better way.

[autumn charm]

It’s nighttime for me and I can’t really tomorrow but thanks for the offer, I hope it’s at least readable

[boreal portal]

It's nighttime for me, too.

[autumn charm]

Oh okay, well I don’t think I really can tomorrow but I’ll see, thank you !

[sand quartz]

Squeeze theorem proof-

Damn

[autumn charm]

you need it for a more rigorous proof?

[autumn charm]

nevermind i probably misunderstood

[sand quartz]

Nah i was impressed
I can try LaTeX ing it properly tmrw if you want

[autumn charm]

it is a pretty neat proof i agree

[autumn charm]

sure do whatev you want you can even make write an epic fantasy story about this proof

[sand quartz]

Is that meant to be an insult ;(

[sand quartz]

Wait @autumn charm could we squeeze theorem like ζ(3) like this by any chance

[sand quartz]

Let $n\geq1\in\bN$, $\theta\notin\pi\bZ$

Using De Moivre’s, we get

$$\sin((2n+1)\theta)$$

$$\implies\Im((\cos\theta +i\sin\theta)^{2n+1}) $$

$$\implies\Im\qty(\sum_{k=0}^{2n+1}\binom{2n+1}{k}i^k\sin^k(\theta)\cos^{2n+1-k}(\theta))$$

$$\implies\sum_{k=0}^n\binom{2n+1}{2k+1}(-1)^k\sin^{2k+1}(\theta)\cos^{2n-2k}(\theta)$$

Now we divide by $\sin^{2n+1}(\theta)$ on each side to get

$$\frac{\sin((2n+1)\theta)}{\sin^{2n+1}(\theta)} = \sum_{k=0}^n(-1)^k\cot^{2(n-k)}(\theta)$$

Now, let us consider the polynomial

$$P(x) = \sum_{k=0}\binom{2n+1}{2k+1}(-1)^kx^{n-k}$$

The roots of this are of the form

$$r \in\qty{\cot^2\qty(\frac{k\pi}{2n+1}):k\leq n\in\bN}$$

Now using the relation between the roots and coefficients of $P$ (Vieta’s formulae), we have that

$$\frac1{2n+1}\binom{2n+1}3 = \sum_{k=1}^n\cot^2\qty(\frac{k\pi}{2n+1})$$

$$\implies \frac{n(2n-1)}3 = \sum_{k=1}^n\cot^2\qty(\frac{k\pi}{2n+1})$$

Now,

$$\sum_{k=1}^n\csc^2\qty(\frac{k\pi}{2n+1}) = \frac{n(2n-1)}3 + n = \frac{2n(n+1)}3$$

Because $\cot^2\theta + 1 = \csc^2\theta$

Now, $\tan(t) \geq t\geq \sin(t)$ $\forall t>0$, this can be proved by convexity of tan and mean value inequality (left as exercise to reader)

So we have that $\forall k\leq n\in\bN$:

$$\tan^2\qty(\frac{k\pi}{2n+1})\geq \frac{k^2\pi^2}{(2n+1)^2}\geq \sin^2\qty(\frac{k\pi}{2n+1})$$

Now, taking the reciprocal and summing up, we get

$$\sum_{k=1}^n\cot^2\qty(\frac{k\pi}{2n+1})\leq \frac{(2n+1)^2}{\pi^2}\sum_{k=1}^n\frac1{k^2}\leq \sum_{k=1}^n\csc^2\qty(\frac{k\pi}{2n+1})$$

$$\implies \frac{n(2n-1)}3 \leq \frac{(2n+1)^2}{\pi^2}\sum_{k=1}^n\frac1{k^2}\leq \frac{2n(n+1}3$$

So finally,

$$\pi^2\frac{n(2n-1)}{3(2n+1)^2} \leq \sum_{k=1}^n \leq \pi^2\frac{2n(n+1)}{3(2n+1)^2}$$

Now both sides tend to $\frac{\pi^2}6$ as $n$ goes to $\infty$

So from the squeeze theorem, we have:

$$\sum_{k=1}^{\infty}\frac1{k^2} = \frac{\pi^2}6$$

Full Credits to Rotor

uwu

[supple lantern]

Ravi #NoLifer

[sand quartz]

@autumn charm this better?

hell I’m saving that

I typed that entire thing on my phone

Only question i have is about the tan(t) ≥ t ≥ sin(t) part

Wouldn’t that be true only for
kπ ≤ t ≤ kπ + π/2 ∀ k ∈ Z

(Ok the bounds are a little stricter but idk)

,w ζ(s-1)/i^s

[autumn charm]

Here it’s true for t in [0;pi/2[ mb

[autumn charm]

Thank you very much, much appreciated

[autumn charm]

No, it was a joke

It wasn’t meant as an insult

I was just saying that you can do whatever you want with it

Im sorry if it was offensive

[sand quartz]

No lmao
Its just that I’m writing a story and a lot of ppl know about it
And some people makes jokes on it

[potent lily]

THIS IS NOT A JOEK

[sand quartz]

Lmao

[autumn charm]

Really? I didn’t know that