Example 3. Determine the radius of convergence of the power series

$$
\sum_{n=0}^{\infty} \cos i n \cdot z^{n}
$$
Solution. We have

$$
c_{n}=\cos i n=\frac{e^{-n}+e^{n}}{2}=\operatorname{ch} n
$$

To find the radius of convergence $R$, we apply formula (6):

$$
\begin{aligned}
R=\lim _{n \rightarrow \infty} \frac{|\operatorname{ch} n|}{|\operatorname{ch}(n+1)|} & =\lim _{n \rightarrow \infty} \frac{\operatorname{ch} n}{\operatorname{ch}(n+1)}=\lim _{n \rightarrow \infty} \frac{\operatorname{ch} n}{\operatorname{ch} n \cdot \operatorname{ch} 1+\operatorname{sh} n \cdot \operatorname{sh} 1}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{\operatorname{ch} 1+\operatorname{th} n \cdot \operatorname{sh} 1}=\frac{1}{\operatorname{ch} 1+\operatorname{sh} 1}=e^{-1}
\end{aligned}
$$

since

$$
\lim _{n \rightarrow \infty} \operatorname{th} n=\lim _{n \rightarrow \infty} \frac{e^{n}+e^{-n}}{e^{n}-e^{-n}}=\lim _{n \rightarrow \infty} \frac{1+e^{-2 n}}{1-e^{-2 n}}=1
$$

Thus, the radius of convergence of the given power series is $R=e^{-1}$,

$D$