How to  solve this :   $$ \prod^{\infty}_{n=2}(1-\frac{1}{n^2}) $$  We  can write it as follows :   $$ \prod^{\infty}_{n=2}\frac{(n-1)(n+1)}{n^2} $$  How can we proceed further please guide.. thanks..
3706
Evaluate the infinite product $$\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).$$  I can't see anything in this limit , so help me please.
52515
I want to find out the following product.  $\prod_0^{n-1}  (r^2 + a^2 -2ra\cos(2k\pi/n - \theta))$ I have been trying to use complex numbers but did not work out. The book I am reading the result is directly stated as $r^{2n} + a^{2n} -2r^na^n\cos(\theta)$. Please help.
1191
how to prove:   $$ \prod_{d|n} d= n^{\frac{\tau (n)}{2}}$$   $\prod_{d|n} d$ is product of all of distinct positive divisor of $n$,  $\tau (n)$ is number (count)of all of positive divisor of $n$
38757
What is   $\prod\limits_{n\geq 2}(1-\frac{1}{n^3})=?$ $\prod\limits_{n\geq 1}(1+\frac{1}{n^3})=?$   I am sure about their convergence. But don't know about exact values. Know some bounds as well. For example first one is in interval (2/3,1) and second one is in (2,3).
54998
While trying some problems along with my friends we had difficulty in this question.   True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.   I couldn't do it and my friend justified it by saying that since the terms in the product have values less than $1$, so the value of the product can never be $1$. I don't know whether this justification is correct or not. But i referred to Tom Apostol's Mathematical Analysis book and found a theorem which states, that    The infinite product $\prod(1-a_{n})$ converges if the series $\sum a_{n}$ converges.   This assures that the above product converges. Could anyone help me in finding out where it converges to? And,   Does there exist a function $f$ in $\mathbb{N}$ ( like $n^{2}$, $n^{3}$) such that $\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)$ has the value $1$?
86635
I try to solve the following sum:  $$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$  I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.
53324
How can I prove that $$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$  I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
18978
Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$  Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
83268
What will be the value of the following Infinite Product :   $$\displaystyle \prod_{k=0}^\infty \left(1+\dfrac{1}{2^{2^k}}\right)$$  It would be nice if anyone could spare the time and boil down to the absolute basics and tell how they reached the solution.
52326
Show that: $$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$  First I will re-define the sum: $$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$  Bow I will define e: $$e^2 = 1+ \frac{2}{1!} + \frac{x^2}{2!} + ... + \infty$$ $$e' = 1 + \frac{1}{1!} + \frac{1}{2!} + ... + \infty$$ $$(e'-2) = \sum^\infty_{n=1} \frac{1}{(n+1)!}$$  Now I need help.
119877
Let $a_0=5/2$ and $a_k=a^2_{k-1}-2$ for all $k\geq 1$.The question is to compute $$\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$  I tried to calculate few terms.$a_0=5/2$, $a_1=17/4,a_2=257/16$ it seems that $a_k$ is of the form $2^{2^k}+2^{-2^{k}}$ however I am not sure about it.How to proceed without doing much guesswork.Any ideas?
71488
How does one evaluate   $ \displaystyle\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $?  Seems fairly straightforward, as I just plugged in some initial values $n = 1, 2, 3,\dotsc$  $n = 1$  $ \sin(y)= 2\sin(\frac{y}{2})\cos(\frac{y}{2})$  $ \cos(\frac{y}{2}) = \sin(y)/2\sin(\frac{y}{2}) $  $n = 2$  $\sin(\frac{y}{2})= 2\sin(\frac{y}{4})\cos(\frac{y}{4}) $  $\cos(\frac{y}{4}) = \sin(\frac{y}{2})/2\sin(\frac{y}{4}) $  Hence,  $ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2\sin(\frac{y}{2}) \times \sin(\frac{y}{2})/2\sin(\frac{y}{4}) ... \times \sin(\frac{y}{2^{n-1}})/2\sin(\frac{y}{2^n})  $   Then some denominators/numerators cancel, reducing the expression into  $ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2^n\sin(\frac{y}{2^n})$  Is this expression no longer reducible?   Because apparently, somehow, the denominator seems to reduce to $y$  How does that work?
107857
The question  Prove that: $$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$    What I've tried  Knowing that: $$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$ evaluating at $z=i$ gives $$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)$$ so: $$ \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}$$  I'm stucked up and don't know how to continue, any help?
62393
Task is to find $$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$$  I tried to make it with double-angle formula: $\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}) = \frac{\prod_{k=0}^\infty \sin(x2^{1-k})}{2^\infty \cdot \prod_{k=0}^\infty \sin(x \cdot 2^{-k})} $ I'm sad to admit, that I'm stuck with that one. So I would appreciate any help, such as pointing the right direction.
114959
So I am stuck on this question and am unsure how to solve it, would be appreciated if someone could help me out, thanks! :)  Question  Consider a sequence $(b_n)_{n=1}^{\infty}$ of non-zero real numbers. By definition, the infinite product $\prod_{n=1}^{\infty}$ converges if the sequence $(p_n)_{n=1}^{\infty}$, where  $p_n=\prod_{k=1}^{n}b_k$  converges to some non-zero number. Prove that the convergence of $\prod_{n=1}^{\infty}b_n$ implies $\lim_{n\to\infty}b_n=1$
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.
59840
How to prove $$ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$ and $$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$