Ytukyg

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?

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
$$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).

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

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 $$left|dfrac zeright|<1$$