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$$\begin{align}\mathsf{\text{Compute}\;\lim_{n\to \infty} \left(a_n\right)^{-1/n^2} \text{where,}\, \\ a_n=\left(1+\frac{1}{n^2}\right)\cdot\left(1+\frac{2^2}{n^2}\right)^2\cdots\left(1+\frac{n^2}{n^2}\right)^{n}}.\end{align}$$

Mathematics
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Jeez ... where do you get these questions??

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Other answers:

Hmm Q37. lol
\[ \huge e^{\frac{-1}{n^2}(\ln((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2....(1+\frac{n^2}{n^2})^n)}\]
Yeah, that's what I did too.
thanks ... i'll save a copy of it!!
\[ \huge e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+\ln(1+\frac{2^2}{n^2})^2+....\ln(1+\frac{n^2}{n^2})^n)}\]
\[ \LARGE e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+2\ln(1+\frac{2^2}{n^2})+....+n\ln(1+\frac{n^2}{n^2}))}\] Looks like same kinda problem we faced before .. there was factorial before ... now we have some other thing!!
try expanding log inside and ... make it more ulgy http://www.wolframalpha.com/input/?i=expand+ln%281%2Bx%29+at+0 .... i'm looosing drive!!!
\[\small-\frac{1}{n^2}\left(\ln\left(n^2+1^2\right) - \ln n^2 + 2\ln\left(n^2 + 2^2\right)-2\ln n^2 + \ldots +n\ln\left(n^2+n^2\right) - n\ln n^2 \right)\]
But I don't think it's of much help.
let's try wolframing!!
lol
i don't know even how to ask wolfram .. lol
Can't we use the inequality here? \[\small \int_1^nx\ln \left(n^2+x^2\right) \le \sum_{x=1}^nx\ln\left(n^2+x^2\right)\le \int_1^nx\ln\left(n^2+x^2\right)+\ln\left(n^2+n^2\right)\]I am not sure, this maybe a fallacy.
Neither do I lol :/
$$\small \int_1^nx\ln \left(n^2+x^2\right) \le \sum_{x=1}^nx\ln\left(n^2+x^2\right)\le \int_1^nx\ln\left(n^2+x^2\right)+n\ln\left(n^2+n^2\right)$$
you sure about this Riemann sum??
I am not :/
Hmm I will come back in an hour or so. I am not on laptop, netbook :/ and this site is lagging alot. Goodluck.
haha ... don't expect anything from me!! if i'm lucky ... then consider yourself lucky!!
I think this works out http://www.wolframalpha.com/input/?i=integrate+x+ln%281%2Bx^2%2F100^2%29+from+0+to+1000 http://www.wolframalpha.com/input/?i=sum+j*ln%281%2Bj^2%2F100^2%29+j%3D1+to+1000
After hell lotta simplification ..... \[ \huge e^{\frac{-1}{n^2}(\frac{n^2}{2} (\ln 4 - 1))} = e^{\frac12 - \ln2} \]

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