sequences and series - How to find the limit $lim limits_{ntoinfty}prod limits_{j=1}^n left( 1+frac{1}{j} left(cos left(frac{tj}{n} right)-1 right) right)$?

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

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 :

$$\exp\left({\int_0^1 \frac{\cos(tx)-x}{x} \, dx}\right) .$$

Answer

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.