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If I apply Cauchy II theorem of limit of sequences to this problem I get limit as $1$ since
$$\lim_{n\to\infty} \frac{( {n!})^\frac{1}{n}}{n} = (\frac{1.2.3..n}{n.n..n})^\frac{1}{n}$$ where $a_1 =\frac{1}{n},a_2=\frac{2}{n}$ and so on. Now I can choose $a_n = \frac{n}{n}$ and using Cauchy's II theorem on limits of sequence, the limit of above problem should be 1. Can somebody explain what I am doing wrong here. The textbook I am referring doesn't put any restriction on $<f_n>$ apart from $<f_n>$ being positive. Is Cauchy theorem even applicable in this case ? Edit - I know it can be solved using Riemann Sums.

deathstroke
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    The denominator is not $n^n$, its $n^1$. – Ty. Jun 18 '20 at 18:03
  • See This Answer. – Mark Viola Jun 18 '20 at 18:28
  • Thanks Mark. I know it can be solved using Riemann sums. But My concern is why Cauchy's 2 theorem doesn't give correct result. – deathstroke Jun 18 '20 at 18:33
  • After the edit by the OP … Note $\lim_{n\to\infty}(a_n)^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ when the limits exist. – Mark Viola Jun 18 '20 at 18:34
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    So, $$\lim_{n\to \infty}\left(\prod_{k=1}^n(k/n)\right)^{1/n}=\lim_{n\to \infty}\frac{\prod_{k=1}^{n+1}(k/(n+1))}{\prod_{k=1}^n(k/n)}=\lim_{n\to \infty} \frac{1}{\left(1+\frac1n\right)^n}=1/e$$ – Mark Viola Jun 18 '20 at 18:47
  • Can you please elaborate? The $\lim_{n \to \infty} \frac{n+1}{n}$ exists. Choose $a_{n+1}= \frac{n+1}{n}$. – deathstroke Jun 18 '20 at 18:50
  • @deathstroke: Cauchy's second limit theorem applies here to the sequence $a_n = \frac{n!}{n^n}$. In this case, $\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!} = \frac{1}{(1+1/n)^n}$. So $\lim_{n \to \infty} a_n^{1/n} = \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = 1/\lim_{n \to \infty}(1+1/n)^n = e^{-1}$. – RRL Jun 18 '20 at 23:56
  • Also $\log x$ is not properly Riemann integrable on $[0,1]$ as it is unbounded. Technically $\frac{1}{n}\sum_{k=1}^n \log(k/n)$ is not a Riemann sum. However, it does converge to the improper integral $\int_0^1 \log x , dx$ as the right-hand sum of a monotone increasing function. The "Riemann sums" corresponding to improper integrals converge only under special circumstances, which luckily arise here. – RRL Jun 19 '20 at 00:05

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