Is there a proof for the following fomula of Euler which does not use complex analysis or fourier analysis? $${\sin (\pi x)} = \pi x\prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2}\right).$$  Suppose that the student studied "Introduction to Real Analysis" by Bartle &amp; Sherbert and they just learned what the convergence of an infinite product means.  Then is there a proof that is suitable for them?
81988
I've been looking at proofs of Euler's sine expansion, that is  $$\frac{\sin x}{x}=\prod_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$$  All the proofs seem to rely on complex analysis and Fourier series.   Is there any more elementary proof?
64569
I have to show that if $|z| &lt; 1$, $z \in \mathbb{C}$, $$(1+z)\prod_{n=1}^{+\infty}(1+z^{2^{n}})= \frac{1}{1-z}$$  I want to understand how to do this kind of exercises, any hint ?
8638
How would I go about finding $\prod_{n=1}^\infty (1+\frac{1}{n^2})$ ? Setting $a_k = \prod_{n=1}^k (1+\frac{1}{n^2})$, the first few partial products are $a_1 = 2$, $a_1 = 2.5$, $a_3 \approx 2.78$, $a_4 \approx 2.591$, $a_{20} \approx 3.50$, and they eventually appear to converge at around $3.639$. (These values were found with a quick python program. So much for my initial conjecture that the product converged to $e$.)   I tried writing $1+\frac{1}{n^2}$ as $\frac{n^2 + 1}{n^2}$, but it wasn't immediately obvious how to prove convergence / find the limit because the product doesn't really factor or telescope. Is there a way to calculate the limit?
30466
I found the following identity when trying to prove something else: $$\sum\limits_{n=1}^\infty\frac{(n-1)!x}{\prod\limits_{k=1}^n (k+x)}\equiv 1,~ \forall x&gt;0$$ I'm sure there's a name for it (or for a more general identity for which this is a special case). Does anyone know the name?
102811
This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $\prod_{n=1}^\infty \cos\left(\large\frac{x}{2^n}\right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that $$\large\prod_{n=1}^\infty\cos\left(\frac{x}{2^n}\right)=\frac{\sin (x)}{x}.$$  I can't figure out how to prove this.
34073
How to prove the following product? $$\frac{\sin(x)}{x}= \left(1+\frac{x}{\pi}\right) \left(1-\frac{x}{\pi}\right) \left(1+\frac{x}{2\pi}\right) \left(1-\frac{x}{2\pi}\right) \left(1+\frac{x}{3\pi}\right) \left(1-\frac{x}{3\pi}\right)\cdots$$
85144
I am stuck trying to prove   $$\prod_{k=1}^{n-1} \sin {\pi k \over n} = {n \over 2^{n-1}}$$  and I'd appreciate help.   What I have done so far:  $z^n - 1 = \prod_{k=1}^n (z - \xi^k)$ where $\xi = e^{2 \pi i k\over n}$. Dividing both sides by $z-1$ we get  $$ {z^n - 1 \over z - 1} = \prod_{k=1}^{n-1} (z - \xi^k)$$  Taking the limit $z \to 1$ on both sides, $$ n = \prod_{k=1}^{n-1} (1 - \xi^k)$$  So I'm getting close to what I want to prove but unfortunately, this is where I am stuck!      How to proceed form here?
8836
Eulers product identity is as follows  \begin{align} \prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 \end{align}  How one can explicitly prove this identity?  Note here $q$ deonotes a complex number satisfying $|q|&lt;1$
24958
I found$\ \prod_{i=1}^\infty \frac{p_i}{p_i-1}\approx 11.0453$ on . Moreover, writing a paper, should one provide a proof or it is trivial? Thanks in advance.
53460
$$f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$$ then how to find limit of $\dfrac{3f(n)}{(n+1)(n+2)}$ as $n\to\infty$? I don't know any series like that. Riemann sum is not working. what to do?
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
88110
I am given the problem: Let $a_n&gt;0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.  We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?
63118
In reference to  about showing that the following interesting series takes on the value  $$\sum_{n=0}^\infty \frac{1}{(2n+1)\operatorname{sinh}((2n+1)\pi)}=\frac{\log(2)}{8}$$  I tried the approach of using the residue theorem, from which one can show that  $$\sum_{n=-\infty}^\infty f(n) = -\sum_k \mathrm{Res}_{z=z_k} [\pi \cot{\pi z} \: f(z)] $$  where $z_k$ are non-integral poles of $\pi \cot{\pi z} f(z)$.  In this case,   $$f(z) = \frac{\mathrm{csch}{[(2 z+1) \pi]}}{2 z+1}$$  There is a pole of order $1$ at $z=-1/2$.  Note that is is not of order $2$ because $\cot{z} \sim (\pi/2) - z$ as $z \rightarrow \pi/2$.  One may then compute the single residue of $\pi \cot{\pi z} f(z)$ as  $$\mathrm{Res}_{z=z_k} \left [\pi \cot{\pi z} \: \frac{\mathrm{csch}{[(2 z+1) \pi]}}{2 z+1} \right ] = -\frac{\pi}{4} $$  So I get as a value for the posted series as $\pi/8$, which is not correct.  Nevertheless, I cannot locate my error.  Perhaps someone can.  For example, the series obviously converges, and quickly.  But is there an error in my assumption that  $$\lim_{N \rightarrow \infty} \oint_{C_N} dz \: \pi \cot{\pi z} \: \frac{\mathrm{csch}{[(2 z+1) \pi]}}{2 z+1} = 0$$  where $C_N$ is a rectaingle symmetric about the real and imaginary axes, extending from $[-N-1/2,N+1/2]$?  Or is there something else I do not see?
32148
I'm having a hard time understanding the following evaluation of the infinite product  $$ \prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right).$$ In particular, I don't understand how you go from line 2 to line 3. Here $\omega = -\frac{1}{2}+ i \frac{\sqrt{3}}{2}$, which is a primitive third root of unity.  $$\begin{align}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)  &amp;=\prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3} \\  &amp;=\lim_{m\to\infty}\frac{1}{m}\prod_{n=2}^m\frac{(n-\omega)(n-\omega^2)}{n^2} \\  &amp;=\lim_{m\to\infty}\frac{\Gamma(m+1-\omega)\Gamma(m+1-\omega^2)}{m(m!)^2\Gamma(-\omega)\Gamma(-\omega^2)(1-\omega)(1-\omega^2)(-\omega)(-\omega^2)} \\  &amp;=\frac{1}{3\Gamma(-\omega)\Gamma(-\omega^2)} \\ &amp;=\frac{\sin{\pi(-\omega)}}{3\pi} \\  &amp;=\frac{\cosh (\frac{\sqrt{3}\pi}{2})}{3\pi } \end{align}$$ EDIT: I think I'm starting to make sense out of this by writing out the terms. $$ \begin{align} \prod_{n=2}^{m} \frac{(n- \omega)(n-\omega^{2})}{n^{2}} &amp;= \frac{(2-\omega)(3-\omega) \cdots (m- \omega)(2-\omega^{2})(3-\omega^{2})\cdots (m- \omega^{2})}{2^{2} \cdot 3^{2} \cdot \cdots  \cdot m^{2}}  \\ &amp;= \frac{1}{(m!)^{2}} \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})} \end{align}$$ Now I have to figure out how to take the limit as $m \to \infty$.
25829
Show that $$\lim_{n \to \infty} \sum\limits_{k=1}^n \frac{n}{n^2+k^2} = \frac{\pi}{4}.$$  I am familiar with Taylor series and Fourier series of the standard functions. I tried to compare with those and see if there is a relation, I don't seen to find any. How do I tackle this?
46536
I need to show that $\dfrac{2}{\pi}&lt;\dfrac{\sin(x)}{x}&lt;1$ for $0&lt;x&lt;\dfrac{\pi}{2}$.  I know I need to use the mean value theorem, would I just say that since $f$ is continuous in the interval (call it I).  We know $\exists c\in I$ such that $f'(c) = \dfrac{f\left(\dfrac{\pi}{2}\right)-f(0)}{\dfrac{\pi}{2}}$. I don't know where this is going though.  Also, would I need to show $\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1$? Is there a way to do this without L'H Rule?
4575
I'm trying to prove  the following claim:     If $f$ is continuous  and $\displaystyle\int_a^\infty |f(x)|\;dx$ is finite then $\lim\limits_{ x \to \infty } f(x)=0$.   Here the counter example of all these sine functions won't do, since  their infinite integral does not converge absolutely.   Any hints for a proof?