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$$ a_n =\left(1+\frac{1^2}{n^2}\right)^1\left(1+\frac{2^2}{n^2}\right)^2\cdots \left( 1 + \frac{n^2}{n^2} \right)^n$$

$$\lim_{n\to\infty} a_n^{-1/n^2}$$

So I tried solving it by taking the logarithm.

Let the limit be $L$.


$$\lim_{n\to\infty}\log(L) = \lim_{n\to\infty} \left(-\frac{1}{n^2}\log\left(1+\frac{1}{n^2}\right)-\frac{2}{n^2}\log\left(1+\frac{2^2}{n^2}\right)-\cdots - \frac{n}{n^2}\log\left(1+\frac{n^2}{n^2}\right)\right).$$

This looks like it should be tractable using the definition of

$$\int^1_0 f(x)\,dx =\lim_{n \to \infty} \frac{1}{n}\sum^n_{r=0}f\left(\frac{r}{n}\right)$$

with taking $f(x) = \log(1+x^2)$ but I am not being able to simplify it to the requisite form. I took this approach mainly because we were recently taught this in school, and it seems to work quite well. I would be interested in alternative solutions too of course. Thanks in advance.

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you sure that this converges? I mean obviously $\left(1+\left(\frac{j}{n}\right)^2\right)^j > 1$. and $\left(1+\left(\frac{n}{n}\right)^2\right)^n=2^n\xrightarrow{n\to\infty} \infty$ – mjb4 Mar 4 '14 at 2:04
@mjb4 Look at the required limit – Michael Tong Mar 4 '14 at 2:06
oh that explains it – mjb4 Mar 4 '14 at 2:07
up vote 5 down vote accepted

I think your try is right. Proceed your work, we have: \begin{equation} \lim_{n\rightarrow\infty}\ln(L)=\lim_{n\rightarrow\infty}\sum_{i=1}^n-\frac{1}{n}\frac{i}{n}\ln\left(1+\left(\frac{i}{n}\right)^2\right)\\ =-\int_0^1x\ln(1+x^2)dx=\frac{1}{2}-\ln2 \end{equation} So $L$ is given by $$L=\frac{e^{1/2}}{2}$$

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Man I made a silly mistake! You're right. :) I was thinking something else, and that just messed up my head. – Sangchul Lee Mar 4 '14 at 2:09
Yeah, I was going to comment that on your post before you deleted it. :p I decided to ask a friend of mine ;) – Sabyasachi Mar 4 '14 at 2:10
@sos440 You are modesty. And thank you for your attention. :) – Lion Mar 4 '14 at 2:11
@Lion not a big deal here though, she just multiplied where she was supposed to subtract, I have made sillier mistakes, so I understand. :p – Sabyasachi Mar 4 '14 at 2:13

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