Is $\sum_{n=2}^∞ (\cos1)^n$ convergent and find its sum Now, I'm going to be clear on this one before anyone complains. Yes, this was part of the lectures, I just wanna know if I've solved this one correctly.
$$\sum_{n=2}^\infty (\cos1)^n$$
So, its a geometric series, where $\cos1$ is just a number raised to $n$ and $|r|<1$, so its convergent.
The sum:
$$S=\frac{a}{1-r}=\frac{(\cos1)^2}{1-\cos1}=0.135$$
 A: Your sum is just a generalized version of the geometric series. 
$$ \sum_{k=0}^{\infty} x^k $$ 
For $ x \ge 1 $ the succession $ a_n \rightarrow + \infty $ so it doesn't go to 0, that follows it diverges. 

For $0<x<1$ we prove by induction that
$$ S_n = \sum_{k=0}^{n} x^k = \frac{1-x^{n+1}}{1-x}$$ 
For $n=0, \space 1= \frac{1-x}{1-x} \quad p_0$ is true 
Suppose n true for $p_n$, we need to control $p_{n+1}$
$S_n + x^{n+1}=\frac{1-x^{n+1}}{1-x}+x^{n+1}=\frac{1-x^{n+2}}{1-x}$ so $p_{n+1}$ is true 

For $x \le 1$ the sum doesn't converge or diverge 

Said that, being $\cos(1)=0.504...<1$ we can say it converge to the result you have found.
A: Result is fine but there is a small typo which seems an issue not totally clear to you.
Recall that
$$S_0=\sum_{n=0}^{\infty}ar^n=\frac a{1-r}$$
and in this case $a=1$ and starting from $n=2$ we obtain
$$S=\sum_{n=2}^{\infty} r^n=\sum_{n=0}^{\infty} r^n -1-r=\frac{1}{1-r}-1-r=\frac{1-(1-r^2)}{1-r}=\frac{r^2}{1-r}=\frac{(\cos1)^2}{1-\cos1}$$

Edit
As an alternative interpretation, as indicated by ArticChar in the comments, we also have
$$S=\sum_{n=2}^{\infty} r^n=r^2\sum_{n=2}^{\infty} r^{n-2}=r^2\sum_{k=0}^{\infty} r^{k}=r^2S_0=\frac{r^2}{1-r}$$
which is another clever way to obtain the result.

Edit 2
Following the previous example, more in general we have
$$S=\sum_{n=n_0}^{\infty} r^n=r^{n_0}\sum_{n=n_0}^{\infty} r^{n-n_0}=r^{n_0}\sum_{k=0}^{\infty} r^{k}=r^{n_0}S_0=\frac{r^{n_0}}{1-r}$$
which explain the recipe given i the video you are referring to, that is to put in the numerator the first value for the series which in your case is indeed $r^2=\cos^2(1)$.
