anonymous
  • anonymous
$$\begin{align}\mathsf{\text{Compute}\;\lim_{n\to \infty} \left(a_n\right)^{-1/n^2} \text{where,}\, \\ a_n=\left(1+\frac{1}{n^2}\right)\cdot\left(1+\frac{2^2}{n^2}\right)^2\cdots\left(1+\frac{n^2}{n^2}\right)^{n}}.\end{align}$$
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Latex isn't processing? Sorry.
anonymous
  • anonymous
Try refreshing.
experimentX
  • experimentX
Jeez ... where do you get these questions??

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More answers

anonymous
  • anonymous
Hmm Q37. lol
experimentX
  • experimentX
\[ \huge e^{\frac{-1}{n^2}(\ln((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2....(1+\frac{n^2}{n^2})^n)}\]
anonymous
  • anonymous
Yeah, that's what I did too.
experimentX
  • experimentX
thanks ... i'll save a copy of it!!
experimentX
  • experimentX
\[ \huge e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+\ln(1+\frac{2^2}{n^2})^2+....\ln(1+\frac{n^2}{n^2})^n)}\]
experimentX
  • experimentX
\[ \LARGE e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+2\ln(1+\frac{2^2}{n^2})+....+n\ln(1+\frac{n^2}{n^2}))}\] Looks like same kinda problem we faced before .. there was factorial before ... now we have some other thing!!
experimentX
  • experimentX
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!!!
anonymous
  • anonymous
\[\small-\frac{1}{n^2}\left(\ln\left(n^2+1^2\right) - \ln n^2 + 2\ln\left(n^2 + 2^2\right)-2\ln n^2 + \ldots +n\ln\left(n^2+n^2\right) - n\ln n^2 \right)\]
anonymous
  • anonymous
But I don't think it's of much help.
experimentX
  • experimentX
let's try wolframing!!
anonymous
  • anonymous
lol
experimentX
  • experimentX
i don't know even how to ask wolfram .. lol
anonymous
  • anonymous
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.
anonymous
  • anonymous
Neither do I lol :/
anonymous
  • anonymous
$$\small \int_1^nx\ln \left(n^2+x^2\right) \le \sum_{x=1}^nx\ln\left(n^2+x^2\right)\le \int_1^nx\ln\left(n^2+x^2\right)+n\ln\left(n^2+n^2\right)$$
experimentX
  • experimentX
you sure about this Riemann sum??
anonymous
  • anonymous
I am not :/
anonymous
  • anonymous
Hmm I will come back in an hour or so. I am not on laptop, netbook :/ and this site is lagging alot. Goodluck.
experimentX
  • experimentX
haha ... don't expect anything from me!! if i'm lucky ... then consider yourself lucky!!
experimentX
  • experimentX
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
experimentX
  • experimentX
After hell lotta simplification ..... \[ \huge e^{\frac{-1}{n^2}(\frac{n^2}{2} (\ln 4 - 1))} = e^{\frac12 - \ln2} \]

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