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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.)

$$$$

enter image description here

enter image description here

enter image description here

enter image description here

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...

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  • $\begingroup$ You got $f(1)$, $f(2)$. $f(k)=1$ for $k\ge 3$. $\endgroup$ – A.S. Mar 12 '16 at 7:31
  • $\begingroup$ @A.S. I didn't got. But Wolfram got them. I have no idea about that...... $\endgroup$ – user143993 Mar 12 '16 at 7:44
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    $\begingroup$ $\log(1+\frac k{n^k})\sim \frac k{n^k}$, hence the sum is $\sim \frac {n^2/2}{n^k}$ and goes to $0$, $\frac 1 2$ or $\infty$ depending on sign of $k-2$. $\endgroup$ – A.S. Mar 12 '16 at 7:52
  • $\begingroup$ @A.S. $$\log(1+\frac k{n^k})\sim \frac k{n^k}$$ → $$\log(1+\frac k{n^2})\sim \frac k{n^2}$$ →$$\sum_{k =1}^n \log{\left(1+ \frac{k}{n^2}\right)} \sim\sum_{k =1}^n \frac{k}{n^2} =\frac{1}{n^2} \cdot \frac{n(n+1)}{2} =\frac{(n+1)}{2n}→\frac{1}{2}$$ Wow $\endgroup$ – user143993 Mar 12 '16 at 8:07
  • $\begingroup$ @A.S. Thanks a lot. At this time, I wonder how can I prove(?) $\log(1+\frac k{n^k})\sim \frac k{n^k}$. $$$$ If $\log(1+\frac k{n})\sim \frac k{n}$, then $\frac{\log{1+x}}{x} → 1$ as $x→\infty$, isn't it?? but this limit goes to zero..$$$$ Please answer for me.$$$$ $\endgroup$ – user143993 Mar 12 '16 at 8:15
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Hint. You may start with $$ 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.

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  • $\begingroup$ Thanks for your answering. By the way, Does 1st inequation come from Taylor expansion? Or any other...? $\endgroup$ – user143993 Mar 12 '16 at 8:14
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    $\begingroup$ @user143993 You are welcome. You may just differentiate each part giving $1-x \leq \frac1{1+x} \leq 1$, which is easy to check, then observing that each part takes the value $0$ for $x=0$ gives the result. Thanks! $\endgroup$ – Olivier Oloa Mar 12 '16 at 8:17
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    $\begingroup$ I checked that inequation, cool~! This IDEA come from Taylor expansion, doesn't this? Really thank you. $\endgroup$ – user143993 Mar 12 '16 at 8:23
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    $\begingroup$ @user143993 The answer is yes! $\endgroup$ – Olivier Oloa Mar 12 '16 at 8:25
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    $\begingroup$ NOW I'm feeling quite relieved! $\endgroup$ – user143993 Mar 12 '16 at 8:30
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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 $k\in\Bbb{N}$.

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  • $\begingroup$ Thanks for your answer. At this time, what does signal $\xi$ mean? $\endgroup$ – user143993 Mar 12 '16 at 8:25
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    $\begingroup$ $0<\xi<x$, it is a number. With Lagrange remainder, you can get the exact estimate to certain degree in Taylor series. $\endgroup$ – Math Wizard Mar 12 '16 at 8:42

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