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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Latex isn't processing? Sorry.

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's what I did too.

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

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

experimentX
 4 years ago
Best 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!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0try 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!!!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But I don't think it's of much help.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0let's try wolframing!!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can'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.

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

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

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

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

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