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.

Could anyone show me how to evaluate this limit?

$$ \lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right) $$

Thanks!

share|improve this question
7  
A limit is evaluated not solved. –  Américo Tavares Jan 26 '13 at 13:48
    
@AméricoTavares Thank you for explanation! –  David Čepelík Jan 26 '13 at 16:41
    
You are welcome! –  Américo Tavares Jan 26 '13 at 16:44
    
See my answer here math.stackexchange.com/a/194880/5418, where you will find more accurate result. –  vesszabo Jan 26 '13 at 18:29
add comment

3 Answers

up vote 5 down vote accepted

Let's see another way $$\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right)=$$ $$\lim_{n\to\infty}en\left(\frac{\left(1+\frac1n\right)^n}{e}-1\right)=$$ $$\lim_{n\to\infty}en\ln\left(\frac{\left(1+\frac1n\right)^n}{e}\right)\times \lim_{n\to\infty}\frac{\left(\displaystyle\frac{\left(1+\frac1n\right)^n}{e}-1\right)}{\ln\left(\displaystyle\frac{\left(1+\frac1n\right)^n}{e}\right)}=$$ $$\lim_{n\to\infty}en\ln\left(\frac{\left(1+\frac1n\right)^n}{e}\right)= $$ $$\lim_{n\to\infty}e \left(n^2\ln\left(1+\frac{1}{n}\right)-n\right)=-\frac{e}{2} $$ where I used $\displaystyle \lim_{x\to1} \frac{x-1}{\ln x}=1$, and then $$\lim_{x\to\infty} \left(x^2\ln\left(1+\frac{1}{x}\right)-x\right)= $$ $$\lim_{y\to0} \left(\frac{1}{y^2}\ln\left(1+y\right)-\frac{1}{y}\right)= $$ $$\lim_{y\to0} \left(\frac{\ln(1+y)-y}{y^2}\right)= $$ that by l'Hôpital's rule turns into $$ \lim_{y\to0} -\frac{1}{2(y+1)}=-\frac{1}{2}$$

Chris.

share|improve this answer
    
Thanks, that's clear! –  David Čepelík Jan 26 '13 at 18:07
    
@David: welcome! :-) –  Chris's sis Jan 26 '13 at 18:35
add comment

Since $\log(1+x)=x-x^2/2+O(x^3)$, we get $$ \begin{align} n\log\left(1+\frac1n\right) &=n\left(\frac1n-\frac1{2n^2}+O\left(\frac1{n^3}\right)\right)\\ &=1-\frac1{2n}+O\left(\frac1{n^2}\right) \end{align} $$ Therefore, since $e^x=1+x+O\left(x^2\right)$, we have $$ \left(1+\frac1n\right)^n=e\left(1-\frac1{2n}+O\left(\frac1{n^2}\right)\right) $$ From here, it is easy to see that $$ n\left(\left(1+\frac1n\right)^n-e\right)=-\frac{e}{2}+O\left(\frac1n\right) $$ Thus, $$ \lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right)=-\frac{e}{2} $$

share|improve this answer
    
In the first equation, it seems to me that the right-hand side of the expression is divided by n, whilst the left-hand side isn't. Is that right or do I miss something? –  David Čepelík Jan 26 '13 at 13:53
    
Sorry, I see now. –  David Čepelík Jan 26 '13 at 13:59
    
I assume you are using Taylor's polynomial to solve this. I am in my first year at college and don't really know what these are about, should have told you before. (It took me a while to realize what it was.) Nonetheless, thank you very much for your help! If you see any other solution, please do let me know. –  David Čepelík Jan 26 '13 at 14:05
    
Yes. The approximations that I use above are Taylor series. This limit might be workable by L'Hospital, but I haven't looked into it. –  robjohn Jan 26 '13 at 14:08
    
Ok, I'll try that. –  David Čepelík Jan 26 '13 at 14:09
add comment

It's possible to do with L'Hopital too but the derivation is a bit lengthy:

$$A = \lim_{n\rightarrow\infty} \frac{\left(1+1/n\right)^n - e}{1/n}$$

Having $0/0$ we can try to use L'Hopital.

Because

$$ \frac{d}{dn} \left(1+1/n\right)^n = \frac{d}{dn} \exp\left(n \ln\left(1+1/n\right)\right) = \left(1+1/n\right)^n \left(\ln\left(1+1/n\right) + \frac{n\left(-1/n^2\right)}{1+1/n}\right) $$

we have

$$A = \lim_{n\rightarrow\infty} \frac{\left(1+1/n\right)^n \left(\ln\left(1+1/n\right) - \frac{1}{n+1}\right)}{-1/n^2} = e \lim_{n\rightarrow\infty} \frac{\left(\ln\left(1+1/n\right) - \frac{1}{n+1}\right)}{-1/n^2}$$

Which is $0/0$ again (Because $\ln(1+\epsilon)\approx \epsilon$ for $\epsilon\ll 1$) , using L'Hopital once more,

$$A = e \lim_{n\rightarrow\infty} \frac{\left(\frac{-1/n^2}{1+1/n} + \frac{1}{(n+1)^2}\right)}{2/n^3} = e \lim_{n\rightarrow\infty} \frac{-n^2}{2(n+1)^2} = -\frac{e}{2} $$

share|improve this answer
add comment

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.