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I'm trying to prove that in the complex plane $\left(1+\frac{z}{n}\right)^n$ converges uniformly to $e^z$ in every closed disc $|z|\leq c$. I thought about showing the sequence as a logarithm of something, but I'm a bit scared because $\operatorname{Log} z$ doesn't behave nicely in the complex plane.

Another idea is expanding the series according to the binom, and exponent according to Taylor. You get that the difference between the functions is the residue of the Taylor sequence, plus: $\sum\limits_{k=0}^n \left| \frac{n^k-\frac{n!}{(n-k)!}}{k!n^k} \right|c^k$. In the nominator you have a polynomial (with respect to $n$) of degree $k-1$, while the polynomial in the denominator is of degree $k$, but I couldn't prove that it tends to zero.

The idea is that because it is a quotient of to polynomials, it behaves like $\frac{1}{n}$, and the sequence $\frac{1}{n} \sum\limits_{k=0}^n \frac{c^k}{k!}$ tends to zero because the sequence converges (to $e^c$).

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closed as off topic by Andrés E. Caicedo, fedja, Michael Renardy, Steven Landsburg, Brendan McKay Jun 2 '12 at 16:54

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A similar question…;. – Davide Giraudo Jun 2 '12 at 14:20
The logarithm is perfectly well behaved as soon as $n>|c|$. – Brendan McKay Jun 2 '12 at 16:54
up vote 0 down vote accepted

I think I have a proof which follows your second idea.

We have $n^k-\frac{n!}{(n-k)!}=\prod_{j=1}^kn-\prod_{j=1}^k(n-j+1)$. We use the following lemma, which can be shown by induction:

Lemma: If $a_j,b_j, 1\leq j\leq N$ are complex numbers of modulus $\leq R$ then $$\left|\prod_{j=1}^Na_j-\prod_{j=1}^Nb_j\right|\leq r^{N-1}\sum_{j=1}^N|a_j-b_j|.$$

We get that $$\left|n^k-\frac{n!}{(n-k)!}\right|\leq n^{k-1}\sum_{j=1}^k(j-1)=n^{k-1}\frac{k(k-1)}2$$ which gives $$\sum_{k=0}^n\left|n^k-\frac{n!}{(n-k)!}\right|\frac 1{n^k}\frac{c^k}{k!}\leq \frac 1n\sum_{k=2}^n\frac{c^k}{(k-2)!}\leq \frac{c^2}ne^c.$$

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That seems to work, thank you! – Dor Jun 2 '12 at 14:45

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