Finding the sum of the geometric series: $$\sum_{n=6}^{\infty} \frac{2}{4^n}$$
My Approach:
$$\sum_{n=6}^{\infty}\frac{2}{4^n}=\sum_{n=0}^{\infty}\frac{2}{4^n}\color{grey}{-\sum_{n=0}^{6}\frac{2}{4^n}}$$
$$=\frac{a}{1-r}-\color{grey}{\frac{a(1-r^n)}{1-r}}$$
Where:
$a=\frac{2}{4^0}=2$,
$r=\frac{1}{4}$,
$n-1=6\implies n=7$
$$\therefore =\frac{2}{1-1/4}-\frac{2(1-(1/4)^7)}{1-1/4}=1.6276 x 10^{-4}$$
This number is obviously too small, and it is obviously incorrect. Because even the first term is $2/4^6$ which is greater than the whole sum calculated.
What am I doing wrong here?
The sum (for the infinite geometric series) is just $a/(1-r)$ where $a$ is "first term" and $r$ is common ratio. There is no need to adjust this, as long as $a,r$ are determined. In your case $a=2/4^6$ and $r=1/4.$