$\sum^\infty_{n=2}{\frac{2^n + (-1)^n}{4^n}}$

Question

How do you find the value of this series?

$$\sum^\infty_{n=2}{\frac{2^n + (-1)^n}{4^n}}$$

I tried writing out the series at $n=2, n=3,$ and $n=4$, and I attempted to look for a pattern with which to take the limit as $n \rightarrow ∞$. However, I couldn't find this pattern.

Are there any other methods by which to find the sum of a series? Is there any rigorous and mathematical method through which to find the $n$th term of the series?

Answer 1

Hints:

*** If $\;|q|<1\;$ , the sum of a geometric series is

$$\sum_{n=0}^\infty q^n=\frac1{1-q}$$

$${}$$

$$***\;\;\;\;\;\sum_{n=2}^\infty\frac{2^n+(-1)^n}{4^n}=\sum_{n=2}^\infty\left(\frac12\right)^n+\sum_{n=2}^\infty\left(-\frac14\right)^n$$

Answer 2

Hint for finite case: if $r\neq 1$ then,

$\sum_{i=0}^{n}r^{i}=1+r+...+r^{n}=\dfrac{1-r^{n+1}}{1-r}$

remember that your sum starts from 2 so, do not forget subtract first 2 terms at the end..