calculus - 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?

Answer

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
$$\sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n=\frac{\frac{25}{4096}}{1-\left(-\frac{5}{64}\right)}=\frac{25}{4416}$$