Evaluating $\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$

Question

Evaluate $$\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$$

This is a geometric series and since $a = \dfrac{1}{2}$ Then the infinite sum is jsut $S = \dfrac{1}{1-\frac{1}{2}} = 2$ Then I multiply by $2$ to get $4$ right? But the actual answer should just be $2$. Am I missing something?

Answer 1

The first term in $\sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2}+\frac{1}{4}+\dots$ is $\frac{1}{2}$, so the sum of the geometric series is $\frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$.

Another way to do it is to absorb the $2$ into the denominator and then change the index from $n$ to $k = n-1$:

$$\sum_{n=1}^{\infty}\frac{2}{2^n} = \sum_{n=1}^{\infty}\frac{1}{2^{n-1}} = \sum_{k=0}^{\infty}\frac{1}{2^{k}} = 1 + \frac{1}{2}+\frac{1}{4}+\dots= \frac{1}{1-\frac{1}{2}} = 2$$

Answer 2

$$\sum\limits_{n = 1}^\infty {\frac{2}{{{2^n}}}} = 2\sum\limits_{n = 1}^\infty {{{\left( {\frac{1} {2}} \right)}^n}} = 2.\frac{{1/2}} {{1 - 1/2}} = 2.1 = 2$$

Answer 3

Almost right. Note that the stated formula works only for

$$\sum_{n=1}^\infty \left(\frac{1}{2}\right)^{n-1} = 2$$