Unlock the Power of Infinite Series: A Simple Formula to Derive Them All!

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Note that the expansion

\[\frac{1}{1-z}=1+z+z^{2}+\cdots+z^{n}+\cdots\quad\mbox{ for }|z|<1\]

can be used to derive expansion for other series. For example:

\[\frac{1}{1+z}{=}\frac{1}{1-(-z)}{=}1{+}(-z){+}(-z)^{2}{+}\cdots{+}(-z)^{n}{+} \cdots=\sum_{n=0}^{\infty}(-1)^{n}z^{n}\mbox{ for }(|z|<1),\]

\[\frac{1}{1-z^{2}}=1+z^{2}+(z^{2})^{2}+\cdots+(z^{2})^{n}+\cdots=\sum_{n=0}^{ \infty}(z^{2})^{n}\mbox{ for }(|z|<1),\]

\[\frac{1}{z-2}=\frac{1}{2\left(\frac{z}{2}-1\right)}=-\frac{1}{2}\cdot\frac{1 }{1-\frac{z}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty}\left(-\frac{z}{2}\right)^{n} =\sum_{n=0}^{\infty}(-1)^{n+1}\frac{z^{n}}{2^{n+1}}\]

\[(\mbox{for }|z/2|<1\mbox{ or }|z|<2).\]

### Differentiating Power Series

There is a vital connection between power series and holomorphic functions, as we illustrate below. To establish \(\sum a_{n}z^{n}\) is differentiable for \(|z|

**Lemma 15**.: _The power series \(\sum a_{n}z^{n}\) and \(\sum na_{n}z^{n-1}\) have the same radius of convergence._

Proof

: Suppose \(\sum|a_{n}z^{n}|\) converges for \(|z|

\[|na_{n}z^{n-1}|=\frac{n}{|z|}\frac{|z|^{n}}{r^{n}}|a_{n}|r^{n}.\]

Since \(|z|/r<1\), the series

\[\sum_{n=0}^{\infty}n\left(\frac{|z|}{r}\right)^{n}\]converges by the ratio test, thus

\[\exists M\quad\text{such that}\quad n\left(\frac{|z|}{r}\right)^{n}\leq M.\]

Hence,

\[|na_{n}z^{n-1}|\leq\frac{M}{|z|}|a_{n}r^{n}|,\]

and now applying the comparison test yields the claim. To show the converse, assume \(\sum na_{n}z^{n-1}\) converges. Then from

\[|a_{n}z^{n}|\leq|z||na_{n}z^{n-1}|\qquad(n\geq 1)\]

the result follows by the comparison test.

The following theorem provides a very important class of holomorphic functions that are quite easy to work with.

**Theorem 125**.: _The power series_

\[f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\]

_defines a holomorphic function in its disc of convergence. The derivative \(f^{\prime}\) of \(f\) is also a power series obtained by differentiating term by term the series for \(f\). Namely_

\[f^{\prime}(z)=\sum_{n=0}^{\infty}na_{n}z^{n-1}.\]

_Furthermore, \(f^{\prime}(z)\) has the same radius of convergence as \(f(z)\)._

Proof

: Let \(f(z)=\sum_{n=1}^{\infty}a_{n}z^{n}\) have radius of convergence \(R>0\), then \(f\) is holomorphic on \(D(0;R)\). By Lemma 15 we know the power series \(\sum a_{n}z^{n}\) and \(\sum na_{n}z^{n-1}\) have the same radius of convergence. Thus we can define

\[g(z):=\sum_{n=1}^{\infty}na_{n}z^{n-1}\quad\text{for}\quad|z|

and we want to show that \(f^{\prime}(z)\) exists and equals \(g(z)\) for \(z\in D(0;R)\). With this in mind for \(z\), \(z_{0}\in D(0;R)\) form

\[\frac{f(z)-f(z_{0})}{z-z_{0}}-g(z_{0})=\sum_{n=1}^{\infty}a_{n}\left(\frac{z^ {n}-z_{0}^{n}}{z-z_{0}}-nz_{0}^{n-1}\right),\]

and show that this tends to \(0\) as \(z\to z_{0}\). We now observe

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**Corollary 26**.: _All the higher derivatives \(f^{\prime},f^{\prime\prime},\ldots,f^{(n)},\ldots\) of the power series \(f(z)=\sum a_{n}z^{n}\) exist for all \(z\) within the disc of convergence and_

\[f^{(k)}(0)=k!a_{k}.\]

Proof

: To see the expression for \(f^{(k)}(0)\), note that if

\[f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}=a_{0}+a_{1}z+a_{2}z^{2}+a_{3}z^{3}+\cdots+a _{n}z^{n}+\cdots\]

then

\[f^{\prime}(z)=a_{1}+2a_{2}z+3a_{3}z^{2}+\cdots+na_{n}z^{n-1}+\cdots\]

\[f^{\prime\prime}(z)=2a_{2}+2\cdot 3a_{3}z^{2}+\cdots+n(n-1)a_{n}z^{n-2}+\cdots\]

and thus

\[f(0)=a_{0},\ f^{\prime}(0)=a_{1},\ f^{\prime\prime}(0)=2a_{2},\ldots,f^{(k)}( 0)=k!a_{k}\]

or equivalently

\[a_{k}=\frac{f^{(k)}(0)}{k!}.\qed\]

**Remark.** This theorem highlights the basic difference between real and complex functions. As you remember, in the real case, one encounters functions which are differentiable \(n\) times, but not \((n+1)\) times. For example, if we take

\[f(x)=\begin{cases}0&\text{if }x\leq 0\\ x^{2}&\text{if }x>0\end{cases},\]

one can see that \(f^{\prime}(0)\) exists but \(f^{\prime\prime}(0)\) does not exist. In the complex case, if the complex derivative \(f^{\prime}\) of \(f\) exists, then all higher derivatives exist.

For another distinction, consider the function \(f\) defined on \(\mathbb{R}\) by:

\[f(x)=\begin{cases}0&\text{for }\ x\leq 0\\ e^{-1/x^{2}}&\text{for }\ x>0\end{cases}.\]

One can prove that \(f\) is infinitely differentiable on \(\mathbb{R}\) and \(f^{(n)}(0)=0\). Thus the Taylor series expansion of this function at \(x=0\) is

\[\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=0\neq f(x).\]