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For a finite case we have $\lim\limits_{n\rightarrow\infty}f(n)\cdot g(n) =\lim\limits_{n\rightarrow\infty}f(n)\cdot\lim\limits_{n\rightarrow\infty}g(n)$ however when is it possible to interchange the folowwing?

$\lim\limits_{n\rightarrow\infty}\prod\limits_{i=1}^n f(i)$

A specific problem I have is $\lim\limits_{n\rightarrow\infty}\prod\limits_{i=1}^n\frac{2i-1}{3i-2}$. I tried taking the natural logarithm and this results in $\lim\limits_{n\rightarrow\infty}\sum\limits_{i=1}^n\ln\left(\frac{2i-1}{3i-2}\right)$.

If I pass the limit then Ill be multiplying 2/3 infinitely often resulting in zero. Is this logic sound? Come to think of it, $\frac{2n-1}{3n-2}<1$ for every $n>1$ so the product is zero. No need to pass the limit. However the original question with $f(n)$ is still interesting.

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  • $\begingroup$ Take the log, and this becomes a question about infinite sums $\endgroup$ – Ben Grossmann Feb 5 '16 at 1:59
  • $\begingroup$ @Omnomnomnom : only on commutative fields.. ? joking, but if $f(n)$ were matrices, I wouldn't be sure of the answer $\endgroup$ – reuns Feb 5 '16 at 2:01
  • $\begingroup$ For a fixed n, the sum interns of log approaches -infinity.... So taking exponentials we approach to zero. As you vary n this is unchanged. So the limit should be zero. I am not sure where you used commutativity here. $\endgroup$ – DBS Feb 5 '16 at 2:11
  • $\begingroup$ I suppose that for $n$ large enough, your individual fraction factors are less than $3/4$, so that your product is dominated by $\lim_n(3/4)^n$, going to zero. Not necessary to bother with logarithms. $\endgroup$ – Lubin Feb 5 '16 at 2:11
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    $\begingroup$ However, this expression is meaningless: $$\lim\limits_{n\rightarrow\infty}\prod\limits_{n=1}^\infty f(n)$$ The inner part means $f(1)f(2)f(3)\dots$, so what does it mean to let $n\to \infty$? $\endgroup$ – Thomas Andrews Feb 5 '16 at 2:29

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