Question

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

Answer 1

Latex isn't processing? Sorry.

Answer 2

Try refreshing.

Answer 3

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

Answer 4

Hmm Q37. lol

Answer 5

\[ \huge e^{\frac{-1}{n^2}(\ln((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2....(1+\frac{n^2}{n^2})^n)}\]

Answer 6

Yeah, that's what I did too.

Answer 7

thanks ... i'll save a copy of it!!

Answer 8

\[ \huge e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+\ln(1+\frac{2^2}{n^2})^2+....\ln(1+\frac{n^2}{n^2})^n)}\]

Answer 9

\[ \LARGE e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+2\ln(1+\frac{2^2}{n^2})+....+n\ln(1+\frac{n^2}{n^2}))}\]

Looks like same kinda problem we faced before .. there was factorial before ... now we have some other thing!!

Answer 10

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!!!

Answer 11

\[\small-\frac{1}{n^2}\left(\ln\left(n^2+1^2\right) - \ln n^2 + 2\ln\left(n^2 + 2^2\right)-2\ln n^2 + \ldots +n\ln\left(n^2+n^2\right) - n\ln n^2 \right)\]

Answer 12

But I don't think it's of much help.

Answer 13

let's try wolframing!!

Answer 14

lol

Answer 15

i don't know even how to ask wolfram .. lol

Answer 16

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.

Answer 17

Neither do I lol :/

Answer 18

$$\small \int_1^nx\ln \left(n^2+x^2\right) \le \sum_{x=1}^nx\ln\left(n^2+x^2\right)\le \int_1^nx\ln\left(n^2+x^2\right)+n\ln\left(n^2+n^2\right)$$

Answer 19

you sure about this Riemann sum??

Answer 20

I am not :/

Answer 21

Hmm I will come back in an hour or so. I am not on laptop, netbook :/ and this site is lagging alot. Goodluck.

Answer 22

haha ... don't expect anything from me!!
if i'm lucky ... then consider yourself lucky!!

Answer 23

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

Answer 24

After hell lotta simplification .....
\[ \huge e^{\frac{-1}{n^2}(\frac{n^2}{2} (\ln 4 - 1))} = e^{\frac12 - \ln2} \]