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

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Ishaan94Best ResponseYou've already chosen the best response.0
Latex isn't processing? Sorry.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Jeez ... where do you get these questions??
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ \huge e^{\frac{1}{n^2}(\ln((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2....(1+\frac{n^2}{n^2})^n)}\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Yeah, that's what I did too.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
thanks ... i'll save a copy of it!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ \huge e^{\frac{1}{n^2}(\ln(1+\frac{1}{n^2})+\ln(1+\frac{2^2}{n^2})^2+....\ln(1+\frac{n^2}{n^2})^n)}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ \LARGE e^{\frac{1}{n^2}(\ln(1+\frac{1}{n^2})+2\ln(1+\frac{2^2}{n^2})+....+n\ln(1+\frac{n^2}{n^2}))}\] Looks like same kinda problem we faced before .. there was factorial before ... now we have some other thing!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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!!!
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
\[\small\frac{1}{n^2}\left(\ln\left(n^2+1^2\right)  \ln n^2 + 2\ln\left(n^2 + 2^2\right)2\ln n^2 + \ldots +n\ln\left(n^2+n^2\right)  n\ln n^2 \right)\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
But I don't think it's of much help.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
let's try wolframing!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
i don't know even how to ask wolfram .. lol
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
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.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
$$\small \int_1^nx\ln \left(n^2+x^2\right) \le \sum_{x=1}^nx\ln\left(n^2+x^2\right)\le \int_1^nx\ln\left(n^2+x^2\right)+n\ln\left(n^2+n^2\right)$$
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
you sure about this Riemann sum??
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Hmm I will come back in an hour or so. I am not on laptop, netbook :/ and this site is lagging alot. Goodluck.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
haha ... don't expect anything from me!! if i'm lucky ... then consider yourself lucky!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
After hell lotta simplification ..... \[ \huge e^{\frac{1}{n^2}(\frac{n^2}{2} (\ln 4  1))} = e^{\frac12  \ln2} \]
 one year ago
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