TITLE: If $z_n \to z$ then $(1+z_n/n)^n \to e^z$
QUESTION [14 upvotes]: We are dealing with $z \in \mathbb{C}$.
I know that
$$
\left(1+ \frac{z}{n} \right)^n \to e^{z}
$$
as $n \to \infty$. So intuitively if $z_n \to z$ then we should have
$$
\left(1+ \frac{z_n}{n} \right)^n \to e^{z}.
$$
If $z_n \in \mathbb{R}$ I would be happy writing
$$
\exp \left(n \log\left(1+ \frac{z_n}{n} \right) \right) = \exp \left(z_n \frac{\log\left(1+ \frac{z_n}{n} \right)-\log(1+0)}{\frac{z_n}{n}-0} \right) \to \exp(z \cdot 1)
$$
where here we are using the definition of derivative.
But if $z \in \mathbb{C}$ the logarithm is multivalued and I'm not sure that the same calculation is allowed.
Is there a different way to show this for complex $z$?

REPLY [5 votes]: You can use the following standard lemma:
Lemma: If $a_{n} $ is a sequence of complex numbers such that $n(a_{n} - 1)\to 0$ then $a_{n} ^{n} \to 1$.
Now let's put $$a_{n} =\dfrac{1+\dfrac{z_{n}}{n}}{1+\dfrac{z}{n}}$$ so that $$n(a_{n} - 1)=\frac{z_{n}-z}{1+z/n}\to 0$$ and then $a_{n} ^{n} \to 1$ and this means that $(1+(z_{n}/n))^{n}$ tends to the same limit as that of $(1+(z/n))^{n}$ and we are done.

The lemma mentioned at the beginning is proved easily by writing $a_{n} =1+b_{n}$ so that $nb_{n} \to 0$ and we have via binomial theorem $$|a_{n} ^{n} - 1|=|(1+b_{n})^{n}-1|=\left|nb_{n}+\frac{n(n-1)}{2!}b_{n}^{2}+\dots\right|$$ and clearly the RHS is bounded above by $$|nb_{n} |+|nb_{n} |^{2}+\cdots = \frac{|nb_{n} |} {1-|nb_{n}|}$$ Since $nb_{n} \to 0$ we are done.