<s>[INST] Find the value of this sum. $$\sum_{n=5}^\infty \frac {2^n-4}{3^n}$$
I tried treating it as a geometric sum but I get 2 as the value, when it should be $\frac {10}{27}$
 [/INST] $\sum=\sum_{n=0}^{\infty} (\frac{2}{3})^n-4\sum_{n=0}^\infty (\frac{1}{3})^n-\sum_{n=0}^4 \frac{2^n-4}{3^n}$
$=\frac{1}{1-\frac{2}{3}}-\frac{4}{1-\frac{1}{3}}+\frac{91}{27}$
$=3-6+\frac{91}{27}$
$=\frac{10}{27}$
QED.
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