Show that the power series $\sum_{n=0}^\infty \frac{z^n}{n!}$ converges uniformly for all $z$.

Question

Show that the power series $\sum_{n=0}^\infty \frac{z^n}{n!}$ converges uniformly for all $z$.

I know that the definition for uniform convergence is: We say ${_}$ converges to uniformly on a subset of Ω iff for every $>0$ there is an $_0∈ℕ$ so that if $>_0$ then $|_()−()|<$ for all $∈$.

BUT I'm not sure if I can use it the same for a power series or if there are different rules with that. Can I? Also, what would the f(z) be for this? I get that $f_n(z)=\frac{z^n}{n!}$, but how do I get $f(z)$?

Answer 1

When considering series of functions such as $z \mapsto \sum_{n=0}^\infty g_n(z)$, you take $f_n$ as the partial sum : $$ f_n(z) = \sum_{k=0}^n g_k(z). $$

However, the exponential series does not converge uniformaly on all $\mathbb C$. It does however uniformly converges on every bounded subset of $\mathbb C$. Indeed, if $|z| \leq M$ :

$$ \left| \frac{z^n}{n!} \right| = \frac{|z|^n}{n!} \leq \frac{M^n}{n!} $$

and the series $\sum_n \frac{M^n}{n!}$ converges (its sum is $e^M$).

Is that clearer?

Answer 2

No, power series have nothing special that would require a different handling.

You have a sequence of functions of $z$ that are the partial sums

$$f_n(z)=\sum_{k=0}^n\frac{z^k}{k!}.$$

Pointwise convergence is verified by the ratio test,

$$\dfrac{\dfrac{z^{k+1}}{(k+1)!}}{\dfrac{z^k}{k!}}=\frac zk<r<1$$ which is true for $k>x$. But as you see, this condition is based a relation between $z$ and $k$ and there cannot be an $n_0$ that works for all $z$.

Answer 3

As has been pointed out, the convergence is not uniform on $\Bbb C$.
Here's how one could show that.

---

Suppose that it was. Then, there exists $N \in \Bbb N$ such that $$\left|\sum_{k=0}^N \dfrac{z^k}{k!} - e^z\right| < 1 \quad (*)$$ for all $z \in \Bbb C$.

We now show that there exists $z_0 \in \Bbb R$ for which $(*)$ is not true.

Note that $e^x \to 0$ as $x\to-\infty$.
Thus, there exists $M < 0$ such that $e^{x} < 1$ for all $x < M$.

On the other hand, note that $\displaystyle\sum_{k=0}^N \dfrac{z^k}{k!}$ is a polynomial and thus, $\left|\displaystyle\sum_{k=0}^N \dfrac{z^k}{k!}\right| \to \infty$ as $z \to -\infty$.

Thus, we can find a real $z_0 < M$ such that $$\left|\sum_{k=0}^N \dfrac{z_0^k}{k!}\right| > 43.$$ For this $z_0$, we also have that $e^{z_0} < 1$. Therefore, we see that

$$\left|\sum_{k=0}^N \dfrac{z_0^k}{k!} - e^{z_0}\right| \ge \left|\sum_{k=0}^N \dfrac{z_0^k}{k!}\right| - |e^{z_0}| > 43 - 1 > 1,$$ giving us the promised contradiction!