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How to show that $$\lim_{n \rightarrow \infty} (\frac{n}{n+a})^n = \frac{1}{e^a}\,\,?$$

If $a=1\,$ then $$(\frac{n}{n+1})^n = (\frac{1}{\frac{n+1}{n}})^n = \frac{1}{(\frac{n+1}{n})^n} \rightarrow \frac{1}{e}$$ as $ n \rightarrow \infty$

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I am a bit confused about what background you have. How do you know that $((n+1)/n)^n \to e$ without knowing (directly) that $((n+1)/n)^n \to 1/e$ ? –  user10676 Oct 30 '12 at 14:05
I don't get it. I know that $(\frac{n+1}{n})^n \rightarrow e$ and I know that $(\frac{n}{n+1})^n \rightarrow 1/e$ –  laovultai Oct 30 '12 at 15:14
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4 Answers

up vote 8 down vote accepted

Hint: Write your function as $\bigg(\frac{n}{n+a}\bigg)^n=\bigg(\frac{1}{1+\frac{a}{n}}\bigg)^n=\frac{1}{(1+\frac{a}{n})^n}$

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This is a nice hint. –  Mhenni Benghorbal Oct 30 '12 at 13:55
but I do not get where it should lead me? –  laovultai Oct 30 '12 at 13:57
@alvoutila: How did you conclude that $$(\frac{n}{n+1})^n = (\frac{1}{\frac{n+1}{n}})^n = \frac{1}{(\frac{n+1}{n})^n} \rightarrow \frac{1}{e}$$ as $ n \rightarrow \infty$? My answer is exactly as you did. In fact $(1+\frac{a}{n})^n$ tends to $\exp(a)$ as $ n \rightarrow \infty$. –  B. S. Oct 30 '12 at 14:05
ok. I thougth that you should use some kind of function like: if $1/f(n) = 1/e$ as $n -> \infty$, then $(1/f(n))^a = 1/e^a$ as $n -> \infty$, –  laovultai Oct 30 '12 at 14:09
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Rewrite your function as $$\displaystyle e ^{-n\ln \left(1+\cfrac an\right)}$$ and do a change of variable $t=\cfrac 1n$ where $t\rightarrow 0$ when $n\rightarrow +\infty$ then apply l'Hopital's.

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Note that - $$\frac{n}{n+a}=\frac{n+a-a}{n+a}=1-\frac{a}{n+a}$$ And therefore for any $a\in\mathbb{R}$: $$\lim_{n\to\infty} \left(1-\frac{a}{n+a}\right)^n=\lim_{n\to\infty} \left(1-\frac{a}{n+a}\right)^{n+a}\cdot\left(1-\frac{a}{n+a}\right)^{-a}=$$$$=\lim_{n\to\infty} \left(1-\frac{a}{n+a}\right)^{n+a}\cdot\lim_{n\to\infty}\left(1-\frac{a}{n+a}\right)^{-a}=e^{-a}$$

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Did I miss something? Why is the down-vote? –  Amihai Zivan Oct 30 '12 at 14:10
I didn't downvoted, but I don't like the fact that $n$ is an integer and $m$ not an integer. Moreover writing $\frac{n}{n+a} = (1 + a/n)^{-1}$ is much simpler. –  user10676 Oct 30 '12 at 14:16
You're right - my mistake. I should have left it with $n+a$... –  Amihai Zivan Oct 30 '12 at 14:18
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for this case we can calculate Ln() of the function and then result is exp() of the answer.

$$ \lim_{n \rightarrow \infty} (\frac{n}{n+a})^n $$ so we have : $$ \text{Ans}= \lim_{n \rightarrow \infty}\left[\ln \left(\frac{n}{n+a}\right)^n \right] =\lim_{n \rightarrow \infty} \left[ n. \ln \left(\frac{n}{n+a}\right) \right] = \infty . 0 $$ $$ \text{Ans}= \lim_{n \rightarrow \infty} \left[ \frac{\ln(\frac{n}{n+a})}{\frac{1}{n}} \right]=\frac{0}{0} \space \rightarrow \space \text{Ambiguity in Mathematics}$$ we can use Hopital's rule,so we have:

$$ \text{Ans}=\lim_{n \rightarrow \infty}\left[ \frac{\frac{a}{n(n+a)}}{\frac{-1}{n^2}} \right]=\lim_{n \rightarrow \infty} \left[ \frac{-an^2}{n(n+a)} \right]=\lim_{n \rightarrow \infty} \left[ \frac{-an}{(n+a)} \right] = -a $$

$$ \lim_{n \rightarrow \infty} (\frac{n}{n+a})^n = e^{\text{Ans}} =e^{-a}=\frac{1}{e^a}$$

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