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


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?

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    $\begingroup$ 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

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.


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