It is desirable to be able to differentiate and multiply power series. Recall
$$
f^{\prime}(x) \equiv \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
$$
Here $h, x, f$ all can have values in $\mathbb{C}$. The following theorem says you can differentiate power series in the most natural way on the disk of convergence, just as you would differentiate a polynomial. This theorem may seem obvious, but it is a serious mistake to think this. You usually cannot differentiate an infinite series whose terms are functions even if the functions are themselves polynomials. The following is special and pertains to power series. It is another example of the interchange of two limits, in this case, the limit involved in taking the derivative and the limit of the sequence of finite sums.
When you formally differentiate a series term by term, the result is called the derived series.

Theorem 8.2.1 Let $\sum_{n=0}^{\infty} a_{n}(x-a)^{n}$ be a Taylor series having radius of convergence $R>0$ and let
$$
f(x) \equiv \sum_{n=0}^{\infty} a_{n}(x-a)^{n}
$$
for $|x-a|<R$. Then
$$
f^{\prime}(x)=\sum_{n=0}^{\infty} a_{n} n(x-a)^{n-1}=\sum_{n=1}^{\infty} a_{n} n(x-a)^{n-1}
$$
and this new differentiated power series, the derived series, has radius of convergence equal to R. Also, $f(x)$ given by the Taylor series is infinitely differentiable on the interior of its disk of convergence.

Proof: First consider the claim that the derived series has radius of convergence equal to $R$. Let $\hat{R}$ be the radius of convergence of the derived series. Then from Proposition $4.10 .13$ and Lemma 8.1.2.
$$
\frac{1}{\hat{R}} \equiv \lim \sup {n \rightarrow \infty}\left|a{n}\right|^{1 / n} n^{1 / n}=\lim \sup {n \rightarrow \infty}\left|a{n}\right|^{1 / n} \equiv \frac{1}{R}
$$
and so $\hat{R}=R$. If $\limsup {n \rightarrow \infty}\left|a{n}\right|^{1 / n}=0$, the same is true of $\limsup {n \rightarrow \infty}\left|a{n}\right|^{1 / n} n^{1 / n}$ and in this case, the series and derived series both have radius of convergence equal to $\infty$.

Now let $r<R$, the radius of convergence of both series, and suppose $|x-a|<r$. Let $\delta$ be small enough that if $|h|<\delta$, then
$$
|x+h-a| \leq|x-a|+|h|<r
$$
also. Thus, $\lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k} r<1$.
Then for $|h|<\delta$, consider the difference quotient.
$$
\frac{f(x+h)-f(x)}{h}=\frac{1}{h} \sum_{k=0}^{\infty} a_{k}\left((x+h-a)^{k}-(x-a)^{k}\right)
$$Using the binomial theorem,
$$
\begin{aligned}
\frac{f(x+h)-f(x)}{h} &=\frac{1}{h} \sum_{k=0}^{\infty} a_{k}\left((x+h-a)^{k}-(x-a)^{k}\right) \
&=\frac{1}{h} \sum_{k=1}^{\infty} a_{k}\left(\sum_{j=0}^{k}\left(\begin{array}{c}
k \
j
\end{array}\right)(x-a)^{j} h^{k-j}-(x-a)^{k}\right) \
&=\sum_{k=1}^{\infty} a_{k}\left(\sum_{j=0}^{k-1}\left(\begin{array}{c}
k \
j
\end{array}\right)(x-a)^{j} h^{(k-1)-j}\right)
\end{aligned}
$$
Then
$$
\begin{aligned}
&\left|\frac{f(x+h)-f(x)}{h}-\sum_{k=1}^{\infty} a_{k} k(x-a)^{k-1}\right| \
=&\left|\sum_{k=1}^{\infty} a_{k}\left(\sum_{j=0}^{k-1}\left(\begin{array}{c}
k \
j
\end{array}\right)(x-a)^{j} h^{(k-1)-j}-k(x-a)^{k-1}\right)\right| \
=&\left|\sum_{k=2}^{\infty} a_{k}\left(\sum_{j=0}^{k-1}\left(\begin{array}{c}
k \
j
\end{array}\right)(x-a)^{j} h^{(k-1)-j}-k(x-a)^{k-1}\right)\right| \
=&\left|\sum_{k=2}^{\infty} a_{k}\left(\sum_{j=0}^{k-2}\left(\begin{array}{c}
k \
j
\end{array}\right)(x-a)^{j} h^{(k-1)-j}\right)\right|
\end{aligned}
$$
Therefore,
$$
\left|\frac{f(x+h)-f(x)}{h}-\sum_{k=1}^{\infty} a_{k} k(x-a)^{k-1}\right| \leq \sum_{k=2}^{\infty}\left|a_{k}\right|\left(\sum_{j=0}^{k-2}\left(\begin{array}{c}
k \
j
\end{array}\right)|x-a|^{j}|h|^{(k-1)-j}\right)
$$
Now it is clear that $k(k-1)\left(\begin{array}{c}k-2 \ j\end{array}\right) \geq\left(\begin{array}{l}k \ j\end{array}\right)$ and so
$$
\begin{aligned}
&=|h| \sum_{k=2}^{\infty}\left|a_{k}\right|\left(\sum_{j=0}^{k-2}\left(\begin{array}{c}
k \
j
\end{array}\right)|x-a|^{j}|h|^{(k-2)-j}\right) \
&\leq|h| \sum_{k=2}^{\infty}\left|a_{k}\right| k(k-1) \sum_{j=0}^{k-2}\left(\begin{array}{c}
k-2 \
j
\end{array}\right)|x-a|^{j}|h|^{(k-2)-j}
\end{aligned}
$$
$=|h| \sum_{k=2}^{\infty}\left|a_{k}\right| k(k-1)(|x-a|+|h|)^{k-2}<|h| \sum_{k=2}^{\infty}\left|a_{k}\right| k(k-1)$ Byrassumption and what was just observed about $\lim {k \rightarrow \infty} k^{1 / k}$ By assumption and what was just observed about $\lim {k \rightarrow \infty} k^{1 / k}$, $$ \lim \sup {k \rightarrow \infty}\left(\left|a{k}\right| k(k-1) r^{k-2}\right)^{1 / k}<1 $$
$$
\left|\frac{f(x+h)-f(x)}{h}-\sum_{k=1}^{\infty} a_{k} k(x-a)^{k-1}\right|<C|h|
$$

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