Calculating a sum of the series $\sum_{n=1}^{\infty} \frac1{2^{n-1}}$

Question

I have the following series:

$\sum_{n=1}^{\infty} \left(1\over2^{n-1}\right) $

I have to calculate its sum. I don't know how to do so. I'd like to get helped. thanks in advance.

Answer 1

Note:

$$\sum_{n=1}^{\infty} \left(1\over2^{n-1}\right) = \sum_{n=0}^{\infty} \left(1\over2^{n}\right) = \sum_{n = 0}^\infty \left(\frac 12\right)^n$$

And so we have a geometric series:

$$ a + ar + ar^2 + ar^3 + ar^4 + \cdots = \sum_{n=0}^\infty \color{blue}{\bf a}\color{red}{\bf r}^n = \frac{\color{blue}{\bf a}}{1-\color{red}{\bf r}} \iff |r| < 1$$

In your case, we have that $a = \color{blue}{\bf 1}$, giving us a sum $\dfrac{1}{1-\color{red}{\bf r}}$, with $\color{red}{\bf r = \dfrac 12} \lt 1$. $$ \sum_{n = 0}^\infty \color{blue}{\bf 1}\cdot \left(\color{red}{\bf \frac 12}\right)^n = \dfrac{1}{1 - \left(\frac 12 \right)} = 2.$$