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## Ishaan94 3 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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1. Ishaan94

Latex isn't processing? Sorry.

2. Ishaan94

Try refreshing.

3. experimentX

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

4. Ishaan94

Hmm Q37. lol

5. experimentX

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

6. Ishaan94

Yeah, that's what I did too.

7. experimentX

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

8. experimentX

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

9. 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!!

10. 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!!!

11. Ishaan94

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

12. Ishaan94

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

13. experimentX

let's try wolframing!!

14. Ishaan94

lol

15. experimentX

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

16. Ishaan94

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.

17. Ishaan94

Neither do I lol :/

18. Ishaan94

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

19. experimentX

you sure about this Riemann sum??

20. Ishaan94

I am not :/

21. Ishaan94

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

22. experimentX

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

23. experimentX
24. 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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