Convergence of differentiated power series

Question

Consider $\displaystyle f(z)=\sum_{k=0}^\infty a_k z^k$ and suppose that $\displaystyle\sum_{n=0}^\infty f^{(n)}(0)$ converges.

Prove that $\forall z \in \mathbb C, \displaystyle\sum_{n=0}^\infty f^{(n)}(z)$ converges.

I'm pretty stumped on this one.

The convergence $\displaystyle\sum_{n=0}^\infty f^{(n)}(0)$ indicates that $\displaystyle\sum_{n=0}^\infty n! a_n$ converges.

What then ?

Any hint is appreciated.

Answer 1

Put $R_n=\sum_{k\geq n} k! a_k$. We have $R_n\to 0$ as $n$ goes to infinity.

Let $N>0$. We have for $n\geq 0$

$$f^{(n)}(z)=\sum_{k\geq n} k! a_k \frac{z^{k-n}}{(k-n)!}=\sum_{l\geq 0} (n+l)! a_{n+l}\frac{z^l}{l!}$$ and hence $$\sum_{n=0}^N f^{(n)}(z)=\sum_{n=0}^N (\sum_{l\geq 0} (n+l)! a_{n+l})\frac{z^l}{l!}$$

We have a finite number of series, so one can exchange the summations:

$$\sum_{n=0}^N f^{(n)}(z)=\sum_{l\geq 0}(\sum_{n=0}^N (n+l)! a_{n+l} ) \frac{z^l}{l!}=\sum_{l\geq 0} (R_l-R_{l+N+1})\frac{z^l}{l!}$$

Hence, if $g(z)=\sum_{l\geq 0}R_l\frac{z^l}{l!}$, we have:

$$\sum_{n=0}^N f^{(n)}(z)-g(z)=-\sum_{l\geq 0}R_{N+l+1}\frac{z^l}{l!}$$

Fix $\varepsilon>0$. As $R_m$ goes to zero as $m\to \infty$, we can find $M$ such that for all $n\geq M$, we have $|R_n|<\varepsilon$. This gives if $N\geq M$:

$$|-\sum_{l\geq 0}R_{N+l+1}\frac{z^l}{l!}|\leq \sum_{l\geq 0}\varepsilon\frac{|z|^l}{l!} =\varepsilon \exp(|z|)$$ and this finish the proof.

Answer 2

My answer is more of a hint than an answer because it applies only if the convergence of $\sum_{n=0}^\infty f^{(n)}(0)$ is assumed to be absolute.

In that case, calculate \begin{eqnarray*} \sum_{n=0}^{\infty}f^{\left(n\right)}\left(z\right) & = & \sum_{n=0}^{\infty}\sum_{\ell=n}^{\infty}a_{\ell}\frac{\ell!}{\left(\ell-n\right)!}\cdot z^{\ell-n}\\ & \overset{\left(\ast\right)}{=} & \sum_{\ell=0}^{\infty}\left[a_{\ell}\cdot\ell!\sum_{n=0}^{\ell}\frac{z^{\ell-n}}{\left(\ell-n\right)!}\right]\\ & \overset{k=\ell-n}{=} & \sum_{\ell=0}^{\infty}\left[a_{\ell}\cdot\ell!\cdot\sum_{k=0}^{\ell}\frac{z^{k}}{k!}\right], \end{eqnarray*} where I used the absolute convergence of $\sum_{\ell=0}^{\infty}a_{\ell}\cdot\ell!$ to justify the interchange of summation in $\left(\ast\right)$ by noting that \begin{eqnarray*} & & \sum_{\ell=0}^{\infty}\left[\left|a_{\ell}\cdot\ell!\right|\cdot\sum_{n=0}^{\ell}\left|\frac{z^{\ell-n}}{\left(\ell-n\right)!}\right|\right]\\ & = & \sum_{\ell=0}^{\infty}\left[\left|a_{\ell}\cdot\ell!\right|\cdot\sum_{k=0}^{\ell}\left|\frac{z^{k}}{k!}\right|\right]\\ & \leq & e^{\left|z\right|}\cdot\sum_{\ell=0}^{\infty}\left|a_{\ell}\cdot\ell!\right|<\infty \end{eqnarray*} This also shows that the final series exists.

EDIT: It might be worth noting that the final series also exists if the series $\sum_{\ell=0}^{\infty}a_{\ell}\ell!$ is only assumed to converge conditionally. To see this, note that \begin{eqnarray*} \sum_{\ell=0}^{\infty}\left[a_{\ell}\ell!\cdot\sum_{k=0}^{\ell}\frac{z^{k}}{k!}\right] & = & e^{z}\cdot\sum_{\ell=0}^{\infty}a_{\ell}\ell!+\sum_{\ell=0}^{\infty}\left[a_{\ell}\ell!\cdot\left(\left(\sum_{k=0}^{\ell}\frac{z^{k}}{k!}\right)-e^{z}\right)\right]\\ & = & e^{z}\cdot\sum_{\ell=0}^{\infty}a_{\ell}\ell!-\sum_{\ell=0}^{\infty}\left[a_{\ell}\ell!\cdot\sum_{k=\ell+1}^{\infty}\frac{z^{k}}{k!}\right].\qquad\left(\square\right) \end{eqnarray*} Now note that $\left(a_{\ell}\cdot\ell!\right)_{\ell\in\mathbb{N}}$ is a null-sequence, hence bounded by some $C>0$. This implies \begin{eqnarray*} \sum_{\ell=0}^{\infty}\left[\left|a_{\ell}\ell!\right|\cdot\sum_{k=\ell+1}^{\infty}\left|\frac{z^{k}}{k!}\right|\right] & \leq & C\cdot\sum_{\ell=0}^{\infty}\sum_{k=\ell+1}^{\infty}\frac{\left|z\right|^{k}}{k!}\\ & = & C\cdot\sum_{k=1}^{\infty}\left[\frac{\left|z\right|^{k}}{k!}\left(\sum_{\ell=0}^{k-1}1\right)\right]\\ & = & C\cdot\left|z\right|\cdot\sum_{k=1}^{\infty}\frac{\left|z\right|^{k-1}}{\left(k-1\right)!}\\ & = & C\cdot\left|z\right|\cdot e^{\left|z\right|}<\infty, \end{eqnarray*} so that the second series in $\left(\square\right)$ even converges absolutely.