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

Question

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?

@Openspace whoops very correct, will try that out just now. Thanks... — jonsno, Dec 07 '17 at 17:32
See related question https://math.stackexchange.com/q/1849697/72031 — Paramanand Singh, Dec 07 '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. — jonsno, Dec 08 '17 at 06:03

robjohn · Accepted Answer · 2017-12-07T18:28:54.347

6

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

edited Dec 07 '17 at 18:28

answered Dec 07 '17 at 18:09

robjohn

345,667

@GuyFsone perhaps you could be more specific about what the answer doesn't explain on how the answer was obtained. – Simply Beautiful Art Dec 07 '17 at 18:15
@GuyFsone: I have added a couple of steps in case that helps explain what it unclear. – robjohn Dec 07 '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 07 '17 at 18:20
@Robjohn thanks, but I cannot understand a basic step, of how log got involved in the second line. – jonsno Dec 07 '17 at 18:20
@samjoe note that $k^{1/x}=e^{(\ln k)/x} $. – Simply Beautiful Art Dec 07 '17 at 18:22
@samjoe: that was explained in the first line. I am using the fact that $e^t=1+t+O!\left(t^2\right)$ and substituting $t=\frac{\log(k)}x.$ Note that $k^{1/x}=e^t$. – robjohn Dec 07 '17 at 18:22
I hope the downvote is not due to a lack of understanding this answer. – Simply Beautiful Art Dec 07 '17 at 18:29
Oops right, didn't even think of that. Nice answer! Thanks a lot! – jonsno Dec 07 '17 at 18:29
@Rob And again how did you split that on line 3? – jonsno Dec 07 '17 at 18:30
@robjohn why was the term in $1/x^2$ dropped ? – Guy Fsone Dec 07 '17 at 18:31
@GuyFsone: it wasn't dropped. That limit was $1$. – robjohn Dec 07 '17 at 18:32
@robjohn then I dont understand that part. it is not clear to me. how you move from first to the second line by pulling out the term $1/x^2$ – Guy Fsone Dec 07 '17 at 18:34
@samjoe: $\left(1+\frac ax\right)\left(1+O!\left(\frac1{x^2}\right)\right)=1+\frac ax+O!\left(\frac1{x^2}\right)$ – robjohn Dec 07 '17 at 18:35
@GuyFsone: see my comment to samjoe above – robjohn Dec 07 '17 at 18:36

Guy Fsone · Answer 2 · 2017-12-08T10:55:20.743

3

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

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

edited Dec 08 '17 at 10:55

answered Dec 07 '17 at 17:35

Guy Fsone

23,903

I don't know why downvotes still exist on your answer, but this answer is also very useful! Thanks a lot! – jonsno Dec 07 '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 07 '17 at 19:47

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

2 Answers2