$8.1$
Functions Defined in Terms of Series
It is time to consider functions other than polynomials. In particular it is time to give a mathematically acceptable definition of functions like $e^{x}, \sin (x)$ and $\cos (x)$. It has been assumed these functions are known from beginning calculus but this is a pretence. Most students who take calculus come through it without a complete understanding of the circular functions. This is because of the reliance on plane geometry in defining them. Fortunately, these functions can be completely understood in terms of power series rather than wretched plane geometry. The exponential function can also be defined in a simple manner using power series. It is tacitly assumed in this presentation that $x \in \mathbb{F}$, either $\mathbb{R}$ or $\mathbb{C}$.
Definition 8.1.1 Let $\left{a_{k}\right}_{k=0}^{\infty}$ be a sequence of numbers. The expression,
$$
\sum_{k=0}^{\infty} a_{k}(x-a)^{k}
$$
is called a Taylor series centered at a. This is also called a power series centered at a. It is understood that $x$ and $a \in \mathbb{F}$, that is, either $\mathbb{C}$ or $\mathbb{R}$.

In the above definition, $x$ is a variable. Thus you can put in various values of $x$ and ask whether the resulting series of numbers converges. Defining $D$ to be the set of all values of $x$ such that the resulting series does converge, define a new function $f$ defined on $D$ having values in $\mathbb{F}$ as $f(x) \equiv \sum_{k=0}^{\infty} a_{k}(x-a)^{k}$. This might be a totally new function, one which has no name. Nevertheless, much can be said about such functions. The following lemma is fundamental in considering the form of $D$ which always turns out to be of the form $B(a, r)$ along with possibly some points $z$ such that $|z-a|=r$. First here is a simple lemma which will be useful.
Lemma 8.1.2 $\lim {n \rightarrow \infty} n^{1 / n}=1$. Proof: It is clear $n^{1 / n} \geq 1$. Let $n^{1 / n}=1+e{n}$ where $0 \leq e_{n}$. Then raising both sides to the $n^{t h}$ power for $n>1$ and using the binomial theorem,
$$
n=\left(1+e_{n}\right)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}
n \
k
\end{array}\right) e_{n}^{k} \geq 1+n e_{n}+(n(n-1) / 2) e_{n}^{2} \geq(n(n-1) / 2) e_{n}^{2}
$$
Thus $0 \leq e_{n}^{2} \leq \frac{n}{n(n-1)}=\frac{1}{n-1}$. From this the desired result follows because $\left|n^{1 / n}-1\right|=$ $e_{n} \leq \frac{1}{\sqrt{n-1}} .$
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Theorem 8.1.3 Let $\sum_{k=0}^{\infty} a_{k}(x-a)^{k}$ be a Taylor series. Then there exists $r \leq \infty$ such that the Taylor series converges absolutely if $|x-a|r$, the Taylor series diverges. If $\lambda<r$ then the Taylor series converges uniformly on the closed disk $|x-a| \leq \lambda$.

Proof: Note $\limsup {k \rightarrow \infty}\left|a{k}(x-a)^{k}\right|^{1 / k}=\limsup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}|x-a|$. Then by the root test, the series converges absolutely if $|x-a| \lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}<1$ and diverges if $|x-a| \limsup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}>1$. Thus define $$ r \equiv\left{\begin{array}{l}1 / \limsup {k \rightarrow \infty}\left|a{k}\right|^{1 / k} \text { if } \infty>\lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}>0 \ \infty \text { if } \lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}=0 \ 0 \text { if } \limsup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}=\infty\end{array}\right. $$
$r \equiv\left{\begin{array}{l}1 / \limsup {k \rightarrow \infty}\left|a{k}\right|^{1 / k} \text { if } \infty>\lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}>0 \ \infty \text { if } \lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}=0 \ 0 \text { if } \limsup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}=\infty\end{array}\right.$ be as described. Then if $|x-a| \leq \lambda$, then $\left|a_{k}(x-a)^{k}\right|^{1 / k}=\lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}|x-a| \leq \lambda \lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k} \leq \frac{\lambda}{r}<\alpha<1$ Next let $\lambda$ be as described. Then if $|x-a| \leq \lambda$, then $$ \lim \sup {k \rightarrow \infty}\left|a{k}(x-a)^{k}\right|^{1 / k}=\lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k}|x-a| \leq \lambda \lim \sup {k \rightarrow \infty}\left|a{k}\right|^{1 / k} \leq \frac{\lambda}{r}<\alpha<1 $$ It follows that for all $k$ large enough and such $x,\left|a_{k}(x-a)^{k}\right|<\alpha^{k}$. Then by the Weierstrass $M$ test, convergence is uniform. Note that the radius of convergence $r$ is given by $\lim \sup _{k \rightarrow \infty}\left|a_{k}\right|^{1 / k} r=1$ Definition 8.1.4 The number in the above theorem is called the radius of convergence and the set on which convergence takes place is called the disc of convergence. Now the theorem was proved using the root test but often you use the ratio test to find the radius of convergence. This kind of thing is typical in math and one must adjust to this fact. The proof of a theorem does not always yield a way to find the thing the theorem speaks about. The above is an existence theorem. There exists a disk of convergence from the above theorem. You find it in specific cases any way that is most convenient. Example 8.1.5 Find the disc of convergence of the Taylor series $\sum_{n=1}^{\infty} \frac{x^{n}}{n}$. Use Corollary 5.4.10. $\lim _{n \rightarrow \infty}\left(\frac{|x|^{n}}{n}\right)^{1 / n}=\lim _{n \rightarrow \infty} \frac{|x|}{\sqrt[n]{n}}=|x|$ because $\lim _{n \rightarrow \infty} \sqrt[n]{n}=$ 1 and so if $|x|<1$ the series converges. The points satisfying $|z|=1$ require special attention. When $x=1$ the series diverges because it reduces to $\sum_{n=1}^{\infty} \frac{1}{n} .$ At $x=-1$ the series converges because it reduces to $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}$ and the alternating series test applies and gives convergence. What of the other numbers $z$ satisfying $|z|=1 ?$ It turns out this series will converge at all these numbers by the Dirichlet test. Example 8.1.6 Find the radius of convergence of $\sum_{n=1}^{\infty} \frac{n^{n}}{n !} x^{n}$. Apply the ratio test. Taking the ratio of the absolute values of the $(n+1)^{t h}$ and the $n^{\text {th }}$ terms $$ \frac{\frac{(n+1)^{(n+1)}}{(n+1) n !}|x|^{n+1}}{\frac{n^{n}}{n !}|x|^{n}}=(n+1)^{n}|x| n^{-n}=|x|\left(1+\frac{1}{n}\right)^{n} \rightarrow|x| e $$ Therefore the series converges absolutely if $|x| e<1$ and diverges if $|x| e>1$. Consequently, $r=1 / e$. This problem assumes that you remember from calculus the last limit. If not, this will be discussed later.

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