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Could anyone show me how to evaluate this limit?

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

Thanks!

Américo Tavares
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David
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3 Answers3

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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, $$ \lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right)=-\frac{e}{2} $$

robjohn
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  • 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 Jan 26 '13 at 13:53
  • Sorry, I see now. – David 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 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 Jan 26 '13 at 14:09
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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.

user 1591719
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4

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,

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

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