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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

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  1. Ishaan94 Group Title
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    Latex isn't processing? Sorry.

    • 2 years ago
  2. Ishaan94 Group Title
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    Try refreshing.

    • 2 years ago
  3. experimentX Group Title
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    Jeez ... where do you get these questions??

    • 2 years ago
  4. Ishaan94 Group Title
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    Hmm Q37. lol

    • 2 years ago
  5. experimentX Group Title
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    \[ \huge e^{\frac{-1}{n^2}(\ln((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2....(1+\frac{n^2}{n^2})^n)}\]

    • 2 years ago
  6. Ishaan94 Group Title
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    Yeah, that's what I did too.

    • 2 years ago
  7. experimentX Group Title
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    thanks ... i'll save a copy of it!!

    • 2 years ago
  8. experimentX Group Title
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    \[ \huge e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+\ln(1+\frac{2^2}{n^2})^2+....\ln(1+\frac{n^2}{n^2})^n)}\]

    • 2 years ago
  9. experimentX Group Title
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    \[ \LARGE e^{\frac{-1}{n^2}(\ln(1+\frac{1}{n^2})+2\ln(1+\frac{2^2}{n^2})+....+n\ln(1+\frac{n^2}{n^2}))}\] Looks like same kinda problem we faced before .. there was factorial before ... now we have some other thing!!

    • 2 years ago
  10. experimentX Group Title
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    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
  11. Ishaan94 Group Title
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    \[\small-\frac{1}{n^2}\left(\ln\left(n^2+1^2\right) - \ln n^2 + 2\ln\left(n^2 + 2^2\right)-2\ln n^2 + \ldots +n\ln\left(n^2+n^2\right) - n\ln n^2 \right)\]

    • 2 years ago
  12. Ishaan94 Group Title
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    But I don't think it's of much help.

    • 2 years ago
  13. experimentX Group Title
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    let's try wolframing!!

    • 2 years ago
  14. Ishaan94 Group Title
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    lol

    • 2 years ago
  15. experimentX Group Title
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    i don't know even how to ask wolfram .. lol

    • 2 years ago
  16. Ishaan94 Group Title
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    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
  17. Ishaan94 Group Title
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    Neither do I lol :/

    • 2 years ago
  18. Ishaan94 Group Title
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    $$\small \int_1^nx\ln \left(n^2+x^2\right) \le \sum_{x=1}^nx\ln\left(n^2+x^2\right)\le \int_1^nx\ln\left(n^2+x^2\right)+n\ln\left(n^2+n^2\right)$$

    • 2 years ago
  19. experimentX Group Title
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    you sure about this Riemann sum??

    • 2 years ago
  20. Ishaan94 Group Title
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    I am not :/

    • 2 years ago
  21. Ishaan94 Group Title
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    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
  22. experimentX Group Title
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    haha ... don't expect anything from me!! if i'm lucky ... then consider yourself lucky!!

    • 2 years ago
  23. experimentX Group Title
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    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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