# How to prove that $\lim \limits_{n \to \infty}(1+a_n)^n=e^p$ when $a_n \to 0$ and $n a_n \to p$?

Assume that $(a_n)$ is a sequence of complex numbers and $p \in \mathbb{C}$. How to prove, maybe without complex logarithm if possible, that:

If $a_n \rightarrow 0$ and $n a_n \rightarrow p$ then $(1+a_n)^n\to e^p$.

Such theorem is used for example in probability in proofs of some limits theorems.

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In your title, you say $na_n\rightarrow p$. But in the body of your question, you say $na_n\rightarrow p$. Which one is the correct one? – Paul Dec 13 '11 at 13:48
Sorry, $n a_n \rightarrow p$. – L.T Dec 13 '11 at 13:52

Start from $$(1+a_n)^n= \sum_{k=0}^{+\infty} c_{k,n}b_{k,n},\quad\text{with}\quad c_{k,n}=[k\leqslant n]\cdot \frac{n!}{n^k(n-k)!},\quad b_{k,n}=\frac{(na_n)^k}{k!}$$ and try to apply Lebesgue dominated convergence theorem. To this end, note that, on the one hand, $|c_{k,n}|\leqslant1$, and on the other hand, $na_n\to p$, hence there exists a finite $A$ such that $|na_n|\leqslant A$ for every $n$, in particular $|b_{k,n}|\leqslant\dfrac{A^k}{k!}$. This yields $$|c_{k,n}b_{k,n}|\leqslant\frac{A^k}{k!}\quad\text{with}\quad \sum_{k=0}^{+\infty}\frac{A^k}{k!}=\mathrm e^A\quad\text{finite},$$ and the conditions of dominated convergence are met. Now, for every fixed $k$, $c_{k,n}\to c_k=1$ and $b_{k,n}\to b_k=\dfrac{p^k}{k!}$ when $n\to\infty$, hence $$\lim\limits_{n\to\infty}(1+a_n)^n=\sum_{k=0}^{+\infty}c_kb_k=\sum_{k=0}^{+\infty}\frac{p^k}{k!}=\mathrm e^p.$$

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The condition that $a_n\to0$ is implied by the condition that $na_n\to p$. – Did Dec 13 '11 at 15:27
Thanks a lot!!! It is a graet answer. – L.T Dec 13 '11 at 15:53

Eited

Too long for a comment, so i edited the first "real" solution.

$$(1+a_n)^n= \sum_{k=0}^n \frac{n!}{k! (n-k)!}a_n^k \,.$$

$$e^{na_n}= \sum_{k=0}^n \frac{n^k}{k!}a_n^k + \sum_{k=n}^\infty \frac{n^k}{k!}a_n^k\,.$$

Thus,

$$\left|(1+a_n)^n- e^{na_n} \right| \leq \left| \sum_{k=0}^n \frac{n(n-1)..(n-k)-n^k}{k!}a_n^k \right| +\left| \sum_{k=n}^\infty \frac{n^k}{k!}a_n^k\right|$$

It is obvious that the second term converges to $0$, you can use the continuity of the expenential+ uniform convergence of the Taylor series.

The first one is

$$\left| \sum_{k=0}^n \frac{(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{k}{n})-1}{k!}(na_n)^k \right|$$, should be eay to show it is convergent to 0.

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$a_n$ is a sequence of complex numbers. Can you use L'Hospital rule in this case? – Paul Dec 13 '11 at 13:55
oh missed that :( – N. S. Dec 13 '11 at 14:08

I think the easiest version (without avoiding complex logarithms) is the following: For $n$ large enough we have $|a_n| < 1$ and thus

$$\begin{eqnarray} (1+a_n)^n &=& \exp(n \log(1+a_n)) \\ &=& \exp\left(n \left(a_n-\sum_{k=2}^\infty (-1)^k\frac{{a_n}^k}{k}\right)\right) \\ &=& \exp(na_n)\exp\left(na_n\left(\sum_{k=1}^\infty (-1)^k \frac{{a_n}^k}{k+1} \right)\right)\\ &\overset{n\to\infty}\longrightarrow& \exp(p)\exp(p\cdot 0) = e^p \end{eqnarray}$$

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