7
$\begingroup$

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

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

$\endgroup$
  • 2
    $\begingroup$ @Davide : If you put that as an answer perhaps you could get a few votes, mine included. =) $\endgroup$ – Patrick Da Silva Nov 13 '11 at 19:46
  • $\begingroup$ 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. $\endgroup$ – DumbQuestion Nov 13 '11 at 20:12
8
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.