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Calculate the limit of $x_n$ as n tends to $\infty$ , where :

$x_n$ = $\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}$

The way I tried calculating the limit :

I've tried using the formula $\lim_{x\to \infty}(1+1/x_n)^(x_n = e$

Firstly I rewrote the expression as $x_n$ = $\sqrt[n]{1-1+(1+1/n)(1+2/n)...(1+n/n)}$ and afterwards I tried writing $x_n$ but I didn't get anywhere ( sorry for no posting the whole math behind the explanation but believe me , I tried but I just can't get the hand of this mathjax language) and even tried to logarithmize the expression but the result was something calculabe only with integrals and I didn't really reach that point with my studies . Lastly I tried using the root test( $\lim_{x\to \infty} \sqrt[n]{x_n} = \lim_{x\to \infty}\frac{x_{n+1}}{x_n}$ ) but i couldn't get to a proper answer . Thank you dearly in advance for your answer !

Serbacul
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    Let me know if the answers to the question How to prove that $\lim \frac{1}{n} \sqrt[n]{(n+1)(n+2)... 2n} = \frac{4}{e}$ also answer your question. – Sarvesh Ravichandran Iyer Jan 17 '22 at 18:04
  • It does , thank you for your response ! Btw is there a way to check if my question is similar to others on this site without buying your time ? – Serbacul Jan 17 '22 at 18:08
  • Everything you need is here. I used Approach0. Since you know mathjax, you can absolutely rifle through half the questions you'd ever have using this tool. It's amazing, and very simple to use. You can use "Raw Query' to search for expressions using MathJax that are not easy to type using the keyboard that you get from the Help menu. – Sarvesh Ravichandran Iyer Jan 17 '22 at 18:09
  • @SarveshRavichandranIyer: the link you mentioned asks a different question than that of the OPs. When using Apraoch0 and any other searcher, make sure that the question outcomes ask the SAME question as the targeted OP. – Mittens Jan 17 '22 at 19:28
  • @OliverDiaz I agree that it looks different at first, but if you take the $\frac 1n$ inside the $n$th root, it goes in as a $\frac 1{n^n}$ . Assigning one $n$ as a denominator to each of $n+1,n+2,\ldots,2n$ leads to $\sqrt[n]{(1+\frac 1n)(1+\frac 2n) \ldots (1+\frac nn)}$. If it's still not obvious, I'm fine with a reopening. Alternative : perhaps this might be better, or even this. – Sarvesh Ravichandran Iyer Jan 17 '22 at 19:31
  • @SarveshRavichandranIyer: that is not the point, the point is that OP is closed using a different question. OF course, using the idea of one problem can lead to a solution to another, but that does not mean that they are the same question (unless this is the Annals of Mathematics and a paper that does provide anything new is rejected) – Mittens Jan 17 '22 at 19:34
  • @OliverDiaz I'll vote to reopen : once this is done, please looked at the linked threads and see if they resemble the question exactly. I think these are better. – Sarvesh Ravichandran Iyer Jan 17 '22 at 19:36

1 Answers1

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$$\log x_n=\frac1n\sum_{k=1}^n\log\left(1+\frac kn\right)$$ is a Riemannian sum of some integrable function over $[0,1]$

$$\log x_\infty=\left.((x+1)(\log(x+1)-x)\right|_0^1=2\log2-1.$$

Mittens
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Woody3
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