Evaluating Summation of $5^{-n}$ from $n=4$ to infinity

Question

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 $\frac1{1-(\frac15)}= \frac54$.

Answer 1

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 $\frac54-5^{-0}-5^{-1}-5^{-2}-5^{-3}=\frac1{500}$.

Answer 2

$$ \sum_{n=4}^\infty\frac{1}{5^n}=\frac{1}{5^4}\sum_{n=4}^\infty\frac{1}{5^{n-4}}=\frac{1}{5^4}\sum_{m=0}^\infty\frac{1}{5^m}=\frac{1}{5^4}\frac{1}{1-1/5}=\frac{1}{500}. $$

Answer 3

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 $$\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)=\frac{1}{500}.$$

Answer 4

$$S_n = (1/5)^4+...+(1/5)^n\ \ \ \ \ (i)$$

$$-(1/5)S_n = -(1/5)^5-...-(1/5)^n-(1/5)^{n+1}\ \ \ \ (ii)$$

$(i)+(ii)$ $$S_n(1-1/5) = (1/5)^4 - (1/5)^{n+1} \Rightarrow (4/5)S_n = 1/625 - (1/5)^{n+1}$$

$\Rightarrow S_n = 1/500 - (5/4)(1/5)^{n+1}$

but

$(1/5)^n \rightarrow 0$