How to evaluate/calculate sums?

Question

If you have this expression,

$$x + x/2 + x/4 + x/8 + \ldots $$, it equates to roughly $2x$, why is this?

Which area of maths is this? To deal with these kind of sums, and getting a result. I am studying Computer science and we get these sequences a lot, but I struggle to evaluate them and make them equal to a smaller version. How can I get better at this?

Thanks in advance.

Answer 1

Rewrite the sum as$$x+\frac 12x+\frac 14x+\frac 18x+\ldots=x\sum\limits_{k\geq0}\left(\frac 12\right)^k$$The inner sum has a closed form. The proof is simple. If we let $S$ denote the result, then$$S=\sum\limits_{k\geq0}r^k=1+r+r^2+\ldots$$$$Sr=\sum\limits_{k\geq0}r^{k+1}=r+r^2+r^3+\ldots$$Subtracting the first equation from the latter gives$$S(1-r)=1\qquad\implies\qquad S=\frac 1{1-r}$$where $|r|<1$ for the series to converge. Therefore$$\begin{align*}x+\frac 12x+\frac 14x+\frac 18x+\ldots=\frac x{1-\frac 12}=2x\end{align*}$$

Answer 2

HINT

We have that

$$x + x/2 + x/4 + x/8 +\ldots+ x/2^n = x(1+1/2+1/4+\ldots+1/2^n)=x\sum_{k=0}^n\left(\frac12\right)^k$$

which is a geometric series.

Answer 3

Notice that the sequence is in Geometric Progression where the number of terms are infinite. I hope you know the formula for finding the sum of series of an infinite GP which is $\frac a {(1-r)}$ where $a$ is the first term and $r$ is the common ratio.