Thursday, 21 June 2018

sequences and series - How to find the limit limlimitsntoinftyprodlimitsnj=1left(1+frac1jleft(cosleft(fractjnright)1right)right)?





lim




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.


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