We're currently analyzing the convergence of function sequences.
I need to prove $$ \lim_{n \to \infty} \left(1 + \frac{z}{n}\right)^n $$
is not uniformly convergent on $\mathbb{C}$. Can I just use the equivalence $\sum_{n=1}^\infty \frac{z^n}{n!}$ and show this is not uniformly convergent to $e^z$? Or is there a way of showing it using the supremum norm?
Thanks in advance.
On
Suppose that the convergence were uniform. Then, for $\epsilon = 1$, there exists $N \in \Bbb N$ such that $$\left|\left(1 + \dfrac{z}{n}\right)^n - e^z\right| < 1$$ for all $n \ge N$ and $z \in \Bbb C.$ (Note that we have used the fact that the sequence converges pointwise to $z\mapsto e^z$.)
In particular, for $z = n$ in the above, we must have the inequality $$e^n - 2^n < 1$$ for all $n \ge N$.
However, note that $e^n - 2^n \to \infty$ as $n \to \infty$ and so, we can find $N_0 \ge N$ such that $$e^{N_0} - 2^{N_0} > 1.$$
This gives us a contradiction.
The point-wise limit is $e^{z}$. If the convergence is uniform then there exists $n_0$ such that $|(1+\frac z n)^{n} -e^{z}| <1$ for all $z$ whenever $n \geq n_0$. Put $z=n$ to get $|2^{n}-e^{n}| <1$ for al $n \geq n_0$. However, $|2^{n}-e^{n}|=e^{n} |1-(\frac 2 e)^{n}| \to \infty$ since $(\frac 2 e)^{n} \to 0$. This contradiction shows that the convergence is not uniform.