I cant calculate this limit correctly...

http://up184.siz.co.il/up2/imzqmoigx0ot.jpg

thank you very much for your help :)

Printable View

- Nov 8th 2009, 11:27 AMUriKlimit
I cant calculate this limit correctly...

http://up184.siz.co.il/up2/imzqmoigx0ot.jpg

thank you very much for your help :) - Nov 8th 2009, 08:57 PMxxlvh
$\displaystyle \lim_{n\to\infty}(1-(\frac{2}{n+2})^{\frac{n+2}{n}}) = \lim_{n\to\infty}(1-(0)^1) = 1$

- Nov 9th 2009, 12:43 AMUriK
thanks for the help but your answer is wrong...

you answered for the limit of one multiplier...but you got n multipliers...so its not that simple...but thanks anyway :) - Nov 9th 2009, 08:19 AMUriK
nevemind I solved it...

the limit is $\displaystyle e^-2$ by the way - Nov 9th 2009, 04:18 PMxxlvh
I realized that but upon inspection, every exponent should go to 0 as n approaches infinity, regardless of the number inside the brackets every multiplier will equal one. Try graphing a few terms, the limit is one. e^-2 is approximately 0.135 it's much too low.

- Nov 9th 2009, 04:53 PMredsoxfan325
I'm inclined to agree. We have here

$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1-\frac{2}{k+2}\right)^{\frac{k+2}{n}}=\lim_{n\to\in fty}\prod_{k=1}^n\left[\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}$ $\displaystyle =\lim_{n\to\infty}\left[\prod_{k=1}^n\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}$

Fact: $\displaystyle \left(1-\frac{1}{\alpha}\right)^{\alpha}$ tends to $\displaystyle e^{-1}$ as $\displaystyle \alpha$ gets large. (In this case $\displaystyle \alpha=\frac{k+2}{2}$.) Now the argument gets a bit informal (but hopefully stays correct), but as $\displaystyle n$ gets very large, the product becomes approximately $\displaystyle e^{-1}\cdot e^{-1}\cdot...e^{-1}=e^{-n}$. So we have:

$\displaystyle \lim_{n\to\infty}\left[\prod_{k=1}^n\left(1-\frac{1}{\frac{k+2}{2}}\right)^{\frac{k+2}{2}}\rig ht]^{\frac{2}{n}}=\lim_{n\to\infty}(e^{-n})^{2/n}=\boxed{e^{-2}}$