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Consider the next power series.

$$\sum_{n=0}^∞ a_nz^n･･･(1)$$

We take the absolute value of $a_n$ and $z$ which make up this power series.

$$\sum_{n=0}^∞ |a_n||z|^n･･･(2)$$

This power series (2) is called absolute convergence if it is finite. Then if the power series (2) converges, the power series (1) also converges.

The convergence radius R refers to the boundary between the area the power series (2) converges absolutely and doesn’t. If $|z|<R$, the power series (1) converges absolutely. Then if $|z|>R$, it diverges. When $|z|=R$, some power series converge, others diverge. Therefore, we need to think about converging or diverging for each power series.

## How to Calculate the Convergence Radius

I’ll show you two methods of finding the convergence radius.

1.The way using an expression with the upper limit as shown below.

$$\limsup_{ n \to \infty } {\sqrt[n]{a_n}}=\frac{1}{R}･･･(3)$$

2.When the limit $\displaystyle \lim_{n\to \infty} \frac{|a_{n+1}|}{|a_n|}$ exists, the following expression holds.

$$\displaystyle \lim_{ n \to \infty }\frac{|a_{n+1}|}{|a_n|}=\frac{1}{R}･･･(4)$$

In both methods, if the limit value is 0, the convergence radius becomes $R=∞$. In other words, if the absolute value $|z|$ is not infinity, the power series always converge.

For example, as shown in the example below, the convergence radius of $e^z$ is $R=∞$. Therefore, if $|z|$ is not infinite, $e^z$ always converges and is a finite value. In fact, if you draw a graph of $e^z$, you can confirm that $e^z$ is a finite value until $z$ goes to infinity.

The first method requires deep knowledge of mathematics and it is troublesome, so this time we will solve the example in the second way.

## Practice

Consider the convergence radius of the next $f(z)$.

\begin{eqnarray} \displaystyle f(z)&=&e^z\\&=&1+z+\frac{z^2}{2!}+･･･\\&=&\sum_{n=0}^N \frac{z^n}{n!}\\&=&\sum_{n=0}^N a_nz^n\end{eqnarray}

$$a_n≡\frac{1}{n!}$$

$$a_{n+1}≡\frac{1}{(n+1)!}.$$
$$R=∞$$