# How to find the limit $\lim \limits_{n\to\infty}\prod \limits_{j=1}^n \left( 1+\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right) \right)$?

$$\lim \limits_{n\to\infty}\prod \limits_{j=1}^n \left( 1+\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right) \right)$$

I am unsure how to find this limit. Are there some general techniques that I should be using when looking for the limit? I know that I should get : $$\exp\left({\int_0^1 \frac{\cos(tx)-x}{x} \, dx}\right) .$$

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@Davide : If you put that as an answer perhaps you could get a few votes, mine included. =) – Patrick Da Silva Nov 13 '11 at 19:46
I deleted the probability tag, it was just that I have seen infinite products as part of probability and was unsure which tags to use. – DumbQuestion Nov 13 '11 at 20:12

Put $a_n:=\prod \limits_{j=1}^n \left( 1+\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right) \right)$. Then $$b_n:=\ln a_n=\sum_{j=1}^n\ln\left(1+\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right)\right).$$ Using the inequality $x-\frac{x^2}2\leq \ln(1+x)\leq x$ for $x>-1$, we get $$\sum_{j=1}^n\frac 1j\left(\cos\left(\frac{tj}n\right)-1\right)-\frac 12\sum_{j=1}^n\frac 1{j^2}\left(\cos\left(\frac{tj}n\right)-1\right)^2\leq b_n\leq \sum_{j=1}^n\frac 1j\left(\cos\left(\frac{tj}n\right)-1\right).$$ Put $c_n:=\sum_{j=1}^n\frac 1j\left(\cos\left(\frac{tj}n\right)-1\right)$. Then $$c_n=\frac1n\sum_{j=1}^n\frac nj\left(\cos\left(\frac{tj}n\right)-1\right)=\frac 1n\sum_{k=1}^nf\left(\frac kn\right),$$ with $f(x)=\frac{\cos (tx)-1}x$, which can be extended by continuity to $\left[0,1\right]$. Hence we get $\lim_{n\to\infty}c_n=\int_0^1f(x)dx=\int_0^1\frac{\cos (tx)-1}xdx$. Now, since $$\sum_{j=1}^n\frac 1{j^2}\left(\cos\left(\frac{tj}n\right)-1\right)^2=\frac 1{n^2}\sum_{j=1}^n\frac {n^2}{j^2}\left(\cos\left(\frac{tj}n\right)-1\right)^2$$ is $\frac 1n$ times a Riemann sum of a continuous function (after extension), it's limit $n\to\infty$ is $0$, and we conclude by the squeeze theorem.

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I prove to limit exist, but i can't find this limit

$$a_n=\lim \limits_{n\to\infty}\prod \limits_{j=1}^n \left[ 1+\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right) \right]$$

$$\to \ln a_n=\lim \limits_{n\to\infty}\sum_{j=1}^n \ln\left[ 1+\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right) \right]$$

Using Taylor's expansion: $$\ln(1+x)=x+O(x)$$

Hence: $$\ln\left[ 1+\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right) \right]=\frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right)+O(\cdots)$$

$$\to \ln a_n=\lim \limits_{n\to\infty}\sum_{j=1}^n \left[ \frac{1}{j} \left(\cos \left(\frac{tj}{n} \right)-1 \right) +O(\cdots)\right] =\int_{0}^{1}\frac{\cos(tx)-1}{x}dx$$

Because: $\int_{0} ^{1} \left(\cos(tx)-1\right) dx\leq 3$

Hence, integral $\int_{0}^{1}\frac{\cos(tx)-1}{x}dx$ is convergent integral

Wolfram alpha

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