Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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$

share|improve this question
    
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

4 Answers 4

up vote 8 down vote accepted

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

share|improve this answer
2  
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$. –  Babak S. Oct 30 '12 at 14:05
1  
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

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.

share|improve this answer

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

share|improve this answer
    
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

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:

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.