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Ishaan94
Group Title
$$\begin{align}\mathsf{\text{Compute}\;\lim_{n\to \infty} \left(a_n\right)^{1/n^2} \text{where,}\, \\ a_n=\left(1+\frac{1}{n^2}\right)\cdot\left(1+\frac{2^2}{n^2}\right)^2\cdots\left(1+\frac{n^2}{n^2}\right)^{n}}.\end{align}$$
 2 years ago
 2 years ago
Ishaan94 Group Title
$$\begin{align}\mathsf{\text{Compute}\;\lim_{n\to \infty} \left(a_n\right)^{1/n^2} \text{where,}\, \\ a_n=\left(1+\frac{1}{n^2}\right)\cdot\left(1+\frac{2^2}{n^2}\right)^2\cdots\left(1+\frac{n^2}{n^2}\right)^{n}}.\end{align}$$
 2 years ago
 2 years ago

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Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Latex isn't processing? Sorry.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Try refreshing.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
Jeez ... where do you get these questions??
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Hmm Q37. lol
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\[ \huge e^{\frac{1}{n^2}(\ln((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2....(1+\frac{n^2}{n^2})^n)}\]
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, that's what I did too.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
thanks ... i'll save a copy of it!!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\[ \huge e^{\frac{1}{n^2}(\ln(1+\frac{1}{n^2})+\ln(1+\frac{2^2}{n^2})^2+....\ln(1+\frac{n^2}{n^2})^n)}\]
 2 years ago

experimentX Group TitleBest 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!!
 2 years ago

experimentX Group TitleBest 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!!!
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
\[\small\frac{1}{n^2}\left(\ln\left(n^2+1^2\right)  \ln n^2 + 2\ln\left(n^2 + 2^2\right)2\ln n^2 + \ldots +n\ln\left(n^2+n^2\right)  n\ln n^2 \right)\]
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
But I don't think it's of much help.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
let's try wolframing!!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
i don't know even how to ask wolfram .. lol
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Neither do I lol :/
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
$$\small \int_1^nx\ln \left(n^2+x^2\right) \le \sum_{x=1}^nx\ln\left(n^2+x^2\right)\le \int_1^nx\ln\left(n^2+x^2\right)+n\ln\left(n^2+n^2\right)$$
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
you sure about this Riemann sum??
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
I am not :/
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
haha ... don't expect anything from me!! if i'm lucky ... then consider yourself lucky!!
 2 years ago

experimentX Group TitleBest 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
 2 years ago

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