Convergence of a summation How do I find out what this summation converges to? I dont even know how I'd start o_o
I assume the first part converges to infinity but dont know how the cos works in this case.
$$\sum^{\infty}_{k=1} 3^{-k}\cos{k}$$
 A: What you have is:
$$
\sum_{k=1}^\infty \frac{\cos k}{3^k}
$$
As Daniel mentioned, Euler's formula will be useful here:
You will notice that $\cos x = \frac{1}{2}(e^{ix}+e^{-ix})$. Thus:
\begin{align}
\sum_{k=1}^\infty \frac{\cos k}{3^k} &= \frac{1}{2}\sum_{k=1}^\infty \frac{e^{ik}+e^{-ik}}{3^k} \\
&= \frac{1}{2}\left(\sum_{k=1}^\infty \frac{e^{ik}}{3^k} + \sum_{k=1}^\infty \frac{e^{-ik}}{3^k}\right) \\
&= \frac{1}{2}\left(\sum_{k=0}^\infty \frac{e^{ik}}{3^k} - 1 + \sum_{k=0}^\infty \frac{e^{-ik}}{3^k} - 1\right) \\
&= \frac{1}{2}\left(\sum_{k=0}^\infty \left(\frac{e^i}{3}\right)^k + \sum_{k=0}^\infty \left(\frac{e^{-i}}{3}\right)^k - 2\right) \\
&= \frac{1}{2}\left(\frac{1}{1-\frac{e^i}{3}} +\frac{1}{1-\frac{e^{-i}}{3}}\right) - 1 \\
&= \frac{1}{2}\left(\frac{3}{3-e^i} +\frac{3}{3-e^{-i}}\right) - 1 \\
\end{align}
Notice that in the geometric series $\left|\frac{e^i}{3}\right| < 1$ and $\left|\frac{e^{-i}}{3}\right| < 1$.
Now, Euler's formula will show that $e^i = \cos(1) + i\sin(1)$ and $e^{-i} = \cos(-1) + i\sin(-1) = \cos(1)-i\sin(1)$. Thus:
\begin{align}
\sum_{k=1}^\infty \frac{\cos k}{3^k} &= \frac{1}{2}\left(\frac{3}{3-e^i} +\frac{3}{3-e^{-i}}\right) - 1 \\
&= \frac{3}{2}\left(\frac{1}{3-\cos(1)-i\sin(1)} +\frac{1}{3-\cos(1)+i\sin(1)}\right) - 1 \\
&= \frac{3}{2}\left(\frac{3-\cos(1)+i\sin(1) + 3-\cos(1)-i\sin(1)}{(3-\cos(1)+i\sin(1))(3-\cos(1)-i\sin(1))}\right) - 1 \\
&= \frac{3}{2}\left(\frac{6-2\cos(1)}{10-6\cos(1)}\right) - 1 \\
&= \frac{9-3\cos(1)}{10-6\cos(1)} - 1 \\
&= \frac{9-3\cos(1) - (10 - 6\cos(1)}{10-6\cos(1))} \\
&= \frac{3\cos(1)-1}{10-6\cos(1)} \\
\end{align}
