Q:

proving convergence of this sequence and calculating limit

So i have this sequence:
$a_{n}=1+ \frac{1}{3}\cos1 +\frac{1}{3^2}\cos2+ ... +\frac{1}{3^n}\cos(n) $
I have to prove it is convergent, and then calculate the limit.
I'm not totally sure how to find the limit of this sequence, so i am stuck at the beginning.
Because of the 2nd task, it's probably not the best idea to try to prove it's a Cauchy's sequence, so i guess it's the best to find the limit(at least the candidate) and then prove that it's convergent with that limit by definition.
But i am stuck at the begining. Thanks in advance.

A:

Hint:
$$
\cos k=\operatorname{Re}\bigl(e^{ik}\bigr)
$$
and
$$
\sum_{k=0}^n\frac{\cos k}{3^k}=\operatorname{Re}\Bigl(\sum_{k=0}^n\Bigl(\frac{e^i}{3}\Bigr)^k\Bigr).
$$
The sum is the sum of a geometric progression of ratio $e^i/3$.
$$
\sum_{k=0}^n\Bigl(\frac{e^i}{3}\Bigr)^k=\frac{(e^i/3)^{n+1}-1}{e^i/3-1}.
$$
Since $|e^i/3|=1/3<1$ we have
$$
\lim_{n\to\infty}\sum_{k=0}^n\Bigl(\frac{e^i}{3}\Bigr)^k=\frac{1}{1-e^i/3}=\frac{3}{3-\cos1-i\sin1}.
$$
The desired limit is the real part, that is
$$
\frac{3(3-\cos1)}{(3-\cos1)^2+\sin^21}.
$$

