Find $\lim\limits_{n\to\infty}{\left(1+\frac{1}{n^k}\right)\left(1+\frac{2}{n^k}\right)\cdots\left(1+\frac{n}{n^k}\right) }$ for $k=1, 2, 3, \cdots$

Question

First of all, I already searched Google, math.stackexchange.com...

I know $$ \lim_{n\rightarrow\infty} \left( 1+ \frac{1}{n} \right) ^n=e$$

That is

$$ \lim_{n\rightarrow\infty} \underbrace{\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\cdots\left(1+\frac{1}{n}\right) }_{\text{n times}} =e$$

$$$$ At this time, I made some problems modifying above.

1. $$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\left(1+\frac{n}{n}\right) } =f(1) $$

2. $$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right)\cdots\left(1+\frac{n}{n^2}\right) } =f(2)$$

3. $$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^3}\right)\left(1+\frac{2}{n^3}\right)\cdots\left(1+\frac{n}{n^3}\right) } =f(3)$$

4. $$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^k}\right)\left(1+\frac{2}{n^k}\right)\cdots\left(1+\frac{n}{n^k}\right) } =f(k)$$

$$$$ After thinking above, I feel I'm spinning my wheels with these limit problems.

Eventually, I searched wolframalpha. And the next images are results of wolfram.

(I take a LOG, because I don't know COMMAND of n-times product.)

$$$$

These result (if we trust wolframalpha) say

$$f(1)=\infty$$ $$f(2)=\sqrt{e}$$ $$f(3)=1$$ $$f(30)=1$$

NOW, I'm asking you for help.

I'd like to know how can I find $f(k)$ (for $k=1,2,3,4, \cdots$ ).

I already used Riemann sum, taking Log... but I didn't get anyhing. ;-(

Thank you for your attention to this matter.

----------- EDIT ---------------------------------

The result for $f(1), f(2), f(3), f(30)$ is an achievement of Wolframalpha, not me.

I'm still spinning my wheel, $f(1), f(2), f(3)$, and so on...

Answer 1

Hint. You may start with $$ x-\frac{x^2}2\leq\log(1+x)\leq x, \quad x\in [0,1], $$ giving, for $n\geq1$, $$ \frac{p}{n^k}-\frac{p^2}{2n^{2k}}\leq\log\left(1+\frac{p}{n^k}\right)\leq \frac{p}{n^k}, \quad 0\leq p\leq n, $$ and $$ \sum_{p=1}^n\frac{p}{n^k}-\sum_{p=1}^n\frac{p^2}{2n^{2k}}\leq \sum_{p=1}^n\log\left(1+\frac{p}{n^k}\right)\leq \sum_{p=1}^n\frac{p}{n^k}, \quad 0\leq p\leq n, $$ or $$ \frac{n(n+1)}{2n^k}-\frac{n(n+1)(2n+1)}{6n^{2k}}\leq \sum_{p=1}^n\log\left(1+\frac{p}{n^k}\right)\leq \frac{n(n+1)}{2n^k} $$ and, for $k\geq3$, as $n \to \infty$, $$ \sum_{p=1}^n\log\left(1+\frac{p}{n^k}\right) \to 0. $$ that is

$$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^k}\right)\left(1+\frac{2}{n^k}\right)\cdots\left(1+\frac{n}{n^k}\right) }=1, \quad k\geq3. $$

The cases $k=1, 2$ are clear.

Answer 2

Edit: By Taylor series with Lagrange remainder $$ \ln(1+x)=x-\frac{x^2}{2(1+\xi)^2} $$ where $0<\xi<x$. So we get $$ x-\frac{x^2}{2}\leqslant \ln(1+x)\leqslant x $$ $$ \frac{n+1}{2}-\frac{(n+1)(2n+1)}{12n}=\sum_{i=1}^{n}\left(\frac{i}{n}-\frac{i^2}{2n^2}\right)\leqslant\sum_{i=1}^{n}\ln(1+\frac{i}{n})\leqslant\sum_{i=1}^{n}\left(\frac{i}{n}\right) =\frac{n+1}{2} $$ Thus $$ \lim_{n\to\infty}\sum_{i=1}^{n}\ln(1+\frac{i}{n})=\infty $$ And $$ \frac{n+1}{2n}-O\left(\frac{1}{n}\right)=\sum_{i=1}^{n}\left(\frac{i}{n^2}-\frac{i^2}{2n^4}\right)\leqslant \sum_{i=1}^{n}\ln(1+\frac{i}{n^2})\leqslant \sum_{i=1}^{n}\left(\frac{i}{n^2}\right)=\frac{n+1}{2n} $$ Hence $$ \lim_{n\to\infty}\sum_{i=1}^{n}\ln(1+\frac{i}{n^2})=\frac1{2} $$ For any $k>2$, since $$ O\left(\frac1{n^{k-2}}\right)-O\left(\frac{1}{n^{2k-3}}\right)=\sum_{i=1}^{n}\left(\frac{i}{n^k}-\frac{i^2}{2n^{2k}}\right)\leqslant \sum_{i=1}^{n}\ln(1+\frac{i}{n^k})\leqslant \sum_{i=1}^{n}\left(\frac{i}{n^k}\right)=O\left(\frac1{n^{k-2}}\right) $$ There is $$ \lim_{n\to\infty}\sum_{i=1}^{n}\ln(1+\frac{i}{n^k})=0 $$

Note: This conclusion holds for $k\in\Bbb{R}$, not only $k\in\Bbb{N}$.