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anonymous
 one year ago
guys, as k>infinity, doesn't (1+1/k)^k tend toward 1, not e?
anonymous
 one year ago
guys, as k>infinity, doesn't (1+1/k)^k tend toward 1, not e?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, nevermind... I was evaluating the limit of 1+1/k only, then raising it to the k power. My bad

phi
 one year ago
Best ResponseYou've already chosen the best response.0In case you are interested in this stuff, here is Khan's videos on compounding and e https://www.khanacademy.org/math/algebra2/exponential_and_logarithmic_func/continuous_compounding/v/introductiontocompoundinterestande

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0\[(1+\frac{1}{k})^k=1+\left(\begin{matrix}k \\ 1\end{matrix}\right)1^{k1}(\frac{1}{k})^1+\left(\begin{matrix}k \\ 2\end{matrix}\right)1^{k2}(\frac{1}{k})^2+...=\\1+\frac{k}{k}+\frac{k(k1)}{2k^2}+...=\\2+\frac{(k1)}{2k}+...\] obviously we have \[2<2+\frac{(k1)}{2k}+...\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0if you try to find upper bound you will find \[2+\frac{(k1)}{2k}+...<3\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0so \[2< (1+\frac{1}{k})^k<\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0in fact \[\lim_{k \rightarrow \infty} (1+\frac{ 1 }{ k })^k \rightarrow e\]
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