$\displaystyle
\sum_{n = 1}^{\infty} \frac {1}{4^n}
$
I tried rewriting it to 4^-n and then using the geometric series test 1/1-r
Which makes it to -1/3 convergence. I am not sure if its correct or not...
Since you have$\displaystyle 1/4^n$, you can rewrite it as $\displaystyle (1/4)^n$. This is because 1 to any power is still 1.
Now, because your indexing starts at 1 instead of 0, we can subtract 1 and then add 1.
This will make the index start at 0. The adding one happens to the exponent so now we have $\displaystyle (1/4)^{n+1}$. This can be written as $\displaystyle (1/4)^1*(1/4)^n$.
Of course, $\displaystyle (1/4)^1=(1/4)$ so we just move that out side the summation.
Thus, you have a geometric series times 1/4 which is your a in the formula $\displaystyle a/(1-r)$