I want to calculate

$ \lim_{n \to \infty }{\frac{{[n(n+1)(n+2)...(2n-1)]}^\frac1n}n} $


$$\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac1n \sum_{k=1}^n f\left(\frac{k}n\right)$$

I know I have to convert $\frac nk$ to $x$, but I am confused since all the factors are multiplied together. Should I use $\log$?


marked as duplicate by YuiTo Cheng, Michael Rozenberg, Lee David Chung Lin, max_zorn, Cesareo Jul 4 at 8:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You should show whatever you have tried so that people might address your problem more accurately. $\endgroup$ – StubbornAtom Mar 6 '17 at 8:26
  • 1
    $\begingroup$ Indeed you should use $\log$. Before that, though, I think you should factor out an $n$ from each term in the numerator. $\endgroup$ – user228113 Mar 6 '17 at 8:26
  • 4
    $\begingroup$ Possible duplicate of Evaluate $\lim_{n \rightarrow \infty} \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$ (The accepted answer is using riemann sum) $\endgroup$ – YuiTo Cheng Jul 4 at 4:14

$$ \begin{align} &\lim_{n\to\infty}\frac{{[n(n+1)(n+2)...(2n-1)]}^{1/n}}n\\ &=\lim_{n\to\infty}{\left[1\left(1+\frac1n\right)\left(1+\frac2n\right)\cdots\left(2-\frac1n\right)\right]}^{1/n}\\ &=\exp\left[\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\log\left(1+\frac kn\right)\right]\\ &=\exp\left[\int_0^1\log(1+x)\,\mathrm{d}x\right]\\ \end{align} $$


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