I want to examine the convergence of $(1-\frac{1}{kn})^n$ for $k\in\mathbb{N}$. My intuition is that for $k=1$, the sequence converges towards $1/e$ and for all other value for $k$, it should converge towards $0$. I'm just unsure about how to prove this.
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3It converges to $e^{-1/k}$. More generally, $\left( 1 + \frac{x}{n} \right)^n$ converges to $e^x$; there are many ways to prove this. – Qiaochu Yuan Nov 21 '17 at 16:29
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How can I show this? – blub Nov 21 '17 at 16:29
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https://math.stackexchange.com/questions/1186210/how-to-show-lim-k-rightarrow-infty-left1-fraczk-rightk-ez?noredirect=1&lq=1 – Guy Fsone Nov 21 '17 at 16:33
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https://math.stackexchange.com/questions/1668114/how-to-find-lim-n-to-infty-left-1-fraczn-rightn?noredirect=1&lq=1 – Guy Fsone Nov 21 '17 at 16:34
3 Answers
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$$\left(1-\frac{1}{kn}\right)^n=\left(\left(1-\frac{1}{kn}\right)^{kn}\right)^{\frac{1}{k}}\to \left(\frac{1}{e}\right)^k$$
tong_nor
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Note the well known limit $$ (1+x/n)^n\to e^x $$ for any $x\in \mathbb{R}$. Take $x=-1/k$.
operatorerror
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$\lim_{n \rightarrow \infty}^{} (1-\frac{1}{kn})^{n}_{}= \lim_{n \rightarrow \infty}^{} e^{\log(1-\frac{1}{kn})^{n}_{}}= e^{ \lim_{n \rightarrow \infty}^{} \frac{\log(1-\frac{1}{kn})}{1/n}}= e^{ \lim_{n \rightarrow 0}^{} \frac{\log(1-\frac{n}{k})}{n}} \stackrel{\text{L'Hospital's rule}}{=} e^{-\frac{1}{k}}_{}$
Sunyam
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