Limit of: $\lim\limits_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}$ [duplicate]

Compute the following limit ($n\in \mathbb{N}$) $$\lim_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}$$

My idea was to use the inequality:

$$\left(\frac{1}{n}\sum_{k=1}^{n} 1^{1/x}\right)^{nx}<\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}<\left(\frac{1}{n}\sum_{k=1}^{n} n^{1/x}\right)^{nx} \\ \implies1<L<n^n$$

This gives that the required limit $L$ lies between $1$ and $n^n$. But how can we find its value?

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• What about considering $e^{\lim \ln f(x)}$ – openspace Dec 7 '17 at 17:31
• @Openspace whoops very correct, will try that out just now. Thanks... – samjoe Dec 7 '17 at 17:32
• See related question math.stackexchange.com/q/1849697/72031 – Paramanand Singh Dec 7 '17 at 20:30
• @Paramanand I think this is almost a duplicate of that. If you think so you can also flag. The answers on that question are also very helpful. – samjoe Dec 8 '17 at 6:03
• Consider it done. – Paramanand Singh Dec 8 '17 at 6:56

Since $e^t=1+t+O\!\left(t^2\right)$, set $t=\frac{\log(k)}x$ \begin{align} \lim_{x\to\infty}\left(\frac1n\sum_{k=1}^nk^{1/x}\right)^{nx} &=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x+O\!\left(\frac1{x^2}\right)\right)^{nx}\\ &=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x\right)^{nx}\lim_{x\to\infty}\left(1+O\!\left(\frac1{x^2}\right)\right)^{nx}\\[3pt] &=\lim_{x\to\infty}\left(1+\frac1n\frac{\log(n!)}x\right)^{nx}\cdot1\\[9pt] &=e^{\log(n!)}\\[15pt] &=n! \end{align}

• @GuyFsone perhaps you could be more specific about what the answer doesn't explain on how the answer was obtained. – Simply Beautiful Art Dec 7 '17 at 18:15
• @GuyFsone: I have added a couple of steps in case that helps explain what it unclear. – robjohn Dec 7 '17 at 18:20
• @GuyFsone why do you insist that must be the answer when the comments below your answer have, multiple times, explained both by example inputting values as well as explaining what you've done wrong? – Simply Beautiful Art Dec 7 '17 at 18:20
• @Robjohn thanks, but I cannot understand a basic step, of how log got involved in the second line. – samjoe Dec 7 '17 at 18:20
• @samjoe note that $k^{1/x}=e^{(\ln k)/x}$. – Simply Beautiful Art Dec 7 '17 at 18:22

Let $$f(h)= \ln\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\right)\implies f'(h) =\frac{\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\ln k \right)}{\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h} \right)}$$ $$f(0)= 0 ~~~and~~~~f'(0)=\left(\frac{1}{n}\sum\limits_{k=1}^{n} \ln k\right)=\color{blue}{ \frac{1}{n}\ln \left(n!\right) }$$

Let $x=1/h$ then we have

$$\lim_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx} =\lim_{h\to 0}\exp\left(\frac{n}{h}\ln\left(\frac{1}{n}\sum\limits_{k=1}^{n} k^{h}\right)\right)\\ =\lim_{h\to 0}\exp\left(n\frac{f(h)}{h}\right) =\color{red}{\exp\left(nf'(0)\right) }=\color{red}{n! }$$

• I don't know why downvotes still exist on your answer, but this answer is also very useful! Thanks a lot! – samjoe Dec 7 '17 at 19:15
• @samjoe your are welcome. the owners of the down votes may not be online anymore may that is why – Guy Fsone Dec 7 '17 at 19:47