Consider the proof of:  $$\zeta(2) = \frac{\pi^{2}}{6}$$  So the proof assume that (because of Euler decomposition) $$\frac{\sin(x)}{x} = \prod_{n &gt; 0}\left(1 - \frac{x^{2}}{(n\pi)^{2}}\right)$$  Why should we consider the $\frac{\sin(x)}{x}$, that $\sin(x)$ has the same decomposition?   Please tell me where's my fault.
24765
As I have heard people did not trust Euler when he first discovered the formula (solution of the ) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler and he gave other proofs.   I believe many of you know some nice proofs of this, can you please share it with us?
35998
I saw this formula on  page 2  $$\pi (n)=\sum_{j=2}^{n}\frac{\sin^{2}\left(\pi \frac{(j-1)!^{2}}{j}\right)}{\sin^{2}(\frac{\pi }{j})}$$  Where $\pi(n)$ is the prime counting function. Is this true? How to prove it?
47861
I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$.  For example, see  or the answers to the question "?"  Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$  I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question).  I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.  I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
113595
Prove that, for a natural number $n$,   $$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$$   This follows from a continued fraction identity for which, I think, there is a lengthy proof. But, I thought, that there may be a direct geometric or another proof. Constructing a polynomial with the sines and tangents roots may be helpful.
39847
How to prove  $$\zeta(2)-\log(x)\log(1-x)=\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)$$ I havent started, any hints?
69968
I am currently reading the book Introduction to Algebraic Number Theory by Apostol. To introduce some important asymptotic formulas, Apostol gives a rough definition of the Riemann zeta function (for $s\in\mathbb{R}^+$),     $$\begin{equation}\zeta(s)=\begin{cases} \sum_{n}n^{-s}, &amp;s&gt;1\\ \lim_{x\to\infty}\left(\sum_{n\leqslant x} \frac{1}{n^s}-\frac{x^{1-s}}{1-s}\right), &amp;0&lt;s&lt;1 \end{cases}\end{equation}$$   The second part really confused me. How could we approach this limit? If we see $\zeta$ as an analytic continuation of $\sum_{n}\frac{1}{n^s}$, it should be written as  $$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1} dx$$ This formula can be easily derived from $\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}dx$ by substituting $x=nu$ (which was exactly what Riemann did in his paper). However, I don't see the connection between this formula and the limit form for $0&lt;s&lt;1$. I am really new to this function so maybe this is a dumb question. But please point it out why we can write $\zeta(s)$ in the limit form for real $0&lt;s&lt;1$.    Also, historically, is the limit form derived from the formal or the converse?    Thanks in advance, any help will be appreciated.
54239
Is it true that for $x\in[0,2\pi]$ we have $$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$$ How can I prove it? For other intervals what is the value of above series if is convergent?
74372
Prove $\sin(x)\tan(x) &gt; x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$  So I did the following: Let $f(x) = \sin(x)\tan(x) - x^2$.  Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) &gt; 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.  I was thinking of approximating it using the following inequality $\sin(x) &lt; x &lt; \tan(x)$.   Is that a good direction?
92471
Is there anybody who can help me show the following?      $$ \sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =0  \qquad\hbox{and}\qquad  \sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0 $$   I can only prove this claim for the sine expression when $k=0$, by grouping the output values such that they all cancel out (for both $n$ even and odd). I have thought about using complex numbers in exponential form to prove it, but I can't do it ... :P   I would really appreciate an algerbraic proof.
111375
I've studied the proof that Euler gave for the famous , and it would seem that while it is technically correct, he does not justify all of his steps properly. Namely, he assumes that   $$\frac{\sin(x)}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\dots$$  simply because they have the same roots, which is really not a strong enough condition. How do you really show that the equality holds?  He then notices that if you use the above equality, and consider it against the Taylor expansion for $\frac{\sin(x)}{x}$, then you can equate the coefficients of the two infinite expansions at each order, and the result of the Basel problem follows. But how do you know that if you have two different expansions for a function, then their coefficients at each order must be equal?  I would really appreciate if someone could show me how to make these two intuitive, yet informal, steps rigorous.
97412
I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$ so maybe I could simplify the above more?
8673
From A Classical Introduction to Modern Number Theory by Ireland and Rosen, page 33:     Corollary 1 (Euler's Theorem). If $(a,m) = 1$, then $a^{\phi(m)} \equiv 1\,(m)$.      Proof. The units in $\mathbb{Z}/m\mathbb{Z}$ form a group of order $\phi(m)$. If $(a,m) = 1$, $\bar{a}$ is a unit. Thus $\bar{a}^{\phi(m)} = \bar{1}$ or $a^{\phi(m)} \equiv 1\,(m)$.   If I'm interpreting this correctly, this proof implicitly uses the fact(?) that if $G$ is a group and $x\in G$, then $x^{|G|}=1$, where $1$ is the identity element of $G$.  If this is indeed true, can someone explain why? (I have had no prior exposure to group theory.)
28253
While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$ Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$  Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$  This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so?
14332
I need to prove that $\dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \sum_{n=1}^\infty a^n \sin(nbx)$ where $|a| &lt; 1$.  It seems that this can be proved by using Euler's formula identities for $\cos(bx)$ and $\sin(bx)$ and substituting $z = e^{ibx}$. From that, I get $$ \dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \dfrac{a (z - 1/z)}{2i(1 - a(z + 1/z) + a^2)}$$ $$= \dfrac{a (z^2 -1)}{2i(z - a(z^2 +1) + za^2)}$$ $$= \dfrac{a (z^2 -1)}{-2ai(z^2 - z(1-a^2)/a + 1)}$$  I don't understand what I am supposed to do next.
93405
Recently, I ran across a product that seems interesting.  Does anyone know how to get to the closed form:  $$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$  I tried using the identity $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$ in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something.  This gives the correct solution if $n$ is odd, but of course evaluates to $0$ if $n$ is even.   So, I tried taking that into account, but must have approached it wrong.   How can this be shown? Thanks everyone.
80866
$\text{What are some interesting cases of $\pi$ appearing in situations that do not seem geometric?}$  Ever since I saw the identity $$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$  and the generalization of $\zeta (2k)$, my perception of $\pi$ has changed. I used to think of it as rather obscure and purely geometric (applying to circles and such), but it seems that is not the case since it pops up in things like this which have no known geometric connection as far as I know. What are some other cases of $\pi$ popping up in unexpected places, and is there an underlying geometric explanation for its appearance?  In other words, what are some examples of $\pi$ popping up in places we wouldn't expect?
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I am hoping to find out where the formula  $$\frac{\pi}{2}=\sum_{k=0}^{\infty}\frac{k!}{\left(2k+1\right)!!}$$ comes from. I can't see how one could begin to prove it.