Evaluating $\sum _{n=2}^{\infty } \frac{(-5)^n}{8^{2 n}}$ making use of geometric series 
Evaluate
  $$\sum _{n=2}^{\infty}\frac{(-5)^n}{8^{2n}}$$
  using geometric series.

I thought it would be possible to split this series such that we have 
$$\sum _{n=2}^{\infty } (-5)^n \cdot \sum _{n=2}^{\infty } \left(\frac{1}{8}\right)^{2 n}$$
However, I am not sure that this is actually possible and I also see that the first sum does not converge, so even if it was possible I am not able to solve it. Could someone walk me through the steps?
 A: HINT: It is $\sum_{n=2}^{\infty}\left(\frac{-5}{64}\right)^n$.
A: Try $$\displaystyle \sum _{n=2}^{\infty } \left(-\frac{5}{64}\right)^n = \frac{25}{4096}\sum _{n=0}^{\infty } \left(-\frac{5}{64}\right)^n$$ as a typical geometric series
A: That's not possible, no. But since $8^{2n} = 64^n$, you can rewrite your series as
$$
\sum \left(\frac{-5}{64}\right)^n,
$$
which should get you on your way. 
A: Hint: $$\sum_{n=2}^{\infty} \left(\frac{-5}{64}\right)^n$$
A: First note that
$$
\sum\limits_{n=2}^{\infty } \frac{(-5)^n}{8^{2 n}}= \sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n
$$
Now let's look at the first two terms of the sum
$$
\left(-\frac{5}{64}\right)^2+ \left(-\frac{5}{64}\right)^3+\dots
$$
$$
=\left(\frac{5}{64}\right)^2- \left(\frac{5}{64}\right)^3+\dots
$$
So now we know that
$$
a= \left(\frac{5}{64}\right)^2=\frac{25}{4096}
$$
And 
$$
r= -\frac{5}{64}
$$
Therefore 
$$
\sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n=\frac{\frac{25}{4096}}{1-\left(-\frac{5}{64}\right)}=\frac{25}{4416}
$$
