Find the convergence radius of series

Question

I need to find the convergence radius of the following series: $$\sum_{n=0}^\infty \cos in \times z^n$$ Since $$\cos in = \cosh n = {\frac{e^{in} + e^{-in}}{2}}$$ therefore, using D'Alembert's property I find the limit as $$\lim {\frac{a_{n+1}}{a_n}} = {\frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} \times z^n \times z \times {\frac{2}{(e^{in} + e^{-in})z^n}} = \lim z \times {\frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$ Therefore the radius should be 1. But the answer on the book is ${\frac{1}{e}}$. What's wrong with my solution?

Answer 1

You have two errors here. The first is that $$ \cosh n = \frac{e^n + e^{-n}}{2} $$ not what you wrote (which is $\cos n$. To remember: $\cosh$ is unbounded on $\mathbb{R}$, so you have "true" exponentials, while $\cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute $$\lim_{n\to\infty} \frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} \tag{1}$$ and not $\lim_{n\to\infty} \frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).

So let's compute (1): $$ \frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} = \frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} = = \frac{e+e^{-(2n+1)}}{1+e^{-2n}} \xrightarrow[n\to\infty]{} \frac{e+0}{1+0} = e\,. $$ Therefore, $$ \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = z\cdot e \tag{2} $$ which explains why the radius is $\frac{1}{e}$.

Answer 2

Hint:

We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence

far more easily,so that we need $$|ez|<1$$ and $$\left|\dfrac ze\right|<1$$