Evaluating Summation of $5^{-n}$ from $n=4$ to infinity The answer is $\frac1{500}$ but I don't understand why that is so. 
I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have $\frac1{1-(\frac15)}= \frac54$.
 A: The problem you have is that you do not know why the formula works to begin with. If you did the situation would be clear. Here's the thing:
$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r} \; \;\;\;\;\; |r|<1.$$
Let $r=\frac{1}{5}$, then you really have the following situation:
$$\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
Let's call this infinite sum $S$ and proceed as follows,
$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
then 
$$ \left(\frac{1}{5}\right)S=\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+\left(\frac{1}{5}\right)^6....\;\;\;\;$$
Subtract the second from the first,
$$S-\left(\frac{1}{5}\right)S=1$$
$$S(1-\left(\frac{1}{5}\right))=1$$
$$S=\frac{1}{1-\left(\frac{1}{5}\right)}$$
$$S=\frac{5}{4}.$$
Now recall what $S$ was and realize,
$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}$$
But, you want the powers to start at $n=4$, so subtract the first 4 powers(0,1,2,3) to get,
$$S-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)$$
which means 
$$\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)=\frac{1}{500}.$$
A: The formula gives from $n=0$ to infinity, but you are asked to sum from $n=4$ to infinity. In this case, you take the terms from $n=0$ to infinity using the formula (which you determined is $\frac54$), and get rid of the extra terms. In this case, we don't need the terms when $n=0,1,2,3$, so we can get rid of those terms by subtracting them. The sum is then $\frac54-5^{-0}-5^{-1}-5^{-2}-5^{-3}=\frac1{500}$.
A: $$
\sum_{n=4}^\infty\frac{1}{5^n}=\frac{1}{5^4}\sum_{n=4}^\infty\frac{1}{5^{n-4}}=\frac{1}{5^4}\sum_{m=0}^\infty\frac{1}{5^m}=\frac{1}{5^4}\frac{1}{1-1/5}=\frac{1}{500}.
$$
A: $$S_n = (1/5)^4+...+(1/5)^n\ \ \ \ \ (i)$$ 
$$-(1/5)S_n = -(1/5)^5-...-(1/5)^n-(1/5)^{n+1}\ \ \ \ (ii)$$
$(i)+(ii)$ $$S_n(1-1/5) = (1/5)^4 - (1/5)^{n+1} \Rightarrow (4/5)S_n = 1/625 - (1/5)^{n+1}$$
$\Rightarrow S_n = 1/500 - (5/4)(1/5)^{n+1}$
but
$(1/5)^n \rightarrow 0$
