2
$\begingroup$

So given a convergent sequence $\{a_n\}_{n=1}^\infty$ with limit $a$, I'd like to prove that

$$\lim_{n\to\infty} \left(1+\frac{a_n}{n}\right)^n=e^a.\quad(1)$$

Knowing that $e$ is defined by

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

the relationship in $(1)$ certainly not unintuitive, and also very useful, but how do I prove it?

In the case of a constant sequence $a_n=k\;\forall n$, it's pretty straightforward as you can write

$$\left(1+\frac{k}{n}\right)^n=\exp\left[\log\left(\left(1+\frac{k}{n}\right)^n\right)\right]=\exp\left[n\log\left(1+\frac{k}{n}\right)\right]=\exp\left[\frac{\log\left(1+\frac{k}{n}\right)}{1/n}\right]$$

and then taking the limit you can apply L'Hôpital's rule to differentiate the numerator and denominator separately and then get the result after a few manipulations. But since $$\frac{\mathrm{d}}{\mathrm{d}x} \log\left(1+\frac{f(x)}{x}\right)=\frac{xf'(x)-f(x)}{xf(x)+x²}$$ I will need to know the derivative $(a_n)'$ with respect to $n$ of the sequence, to use this approach, which is not necessarily well-defined. Is there another way to go about this?

$\endgroup$
  • 1
    $\begingroup$ You don't need $a_n'$. In the expression $n\log(1 + a_n/n) = a_n\frac{\log(1 + a_n/n)}{a_n/n}$ when taking the limit $n\to \infty$ we have $a_n/n \to 0$ so you get the same limit $\lim_{x\to 0} \log(1+x)/x$ as you have in the $a_n = k$ case. The only new result you need to use is the basic limit result: if $A_n\to A$ and $B_n\to B$ then $A_nB_n \to AB$. $\endgroup$ – Winther Apr 20 '15 at 1:07
  • $\begingroup$ You're nearly there. Since $a_n\rightarrow a$, for any $\epsilon>0$, for $n$ sufficiently large, $a-\epsilon<a_n<a+\epsilon$, so, you can squeeze between these by letting $\epsilon$ go to zero (perhaps replacing the limits with $\limsup$ and $\liminf$). $\endgroup$ – Michael Burr Apr 20 '15 at 1:10
  • $\begingroup$ Right it all makes sense now. Thanks a lot you guys :-) $\endgroup$ – Erik Olesen Apr 20 '15 at 1:44
1
$\begingroup$

Nothing new here, but perhaps streamlined a bit: Apply $\ln$ to get $$(1)\,\,\,\,n\ln (1+a_n/n)=\frac{\ln (1+a_n/n)}{a_n/n}\cdot a_n.$$ Now as $h\to 0,$ $\ln(1+h)/h = (\ln(1+h)-\ln 1)/h \to \ln'(1) = 1\,$ by definition of the derivative. It follows that the limit in (1) is $a.$ Exponentiating back gives the limit of $e^a$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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