# Can an infinite series have a sum?

Jul 31, 2015

Yes.

For example $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots = {\sum}_{n = 0}^{\infty} {2}^{-} n = 2$

#### Explanation:

In general, if $r \in \left(- 1 , 1\right)$ and $a \ne 0$, then:

${\sum}_{n = 0}^{\infty} a {r}^{n} = \frac{a}{1 - r}$

To see this notice that:

$\left(1 - r\right) {\sum}_{n = 0}^{\infty} a {r}^{n} = {\sum}_{n = 0}^{\infty} a {r}^{n} - r {\sum}_{n = 0}^{\infty} a {r}^{n}$

$= {\sum}_{n = 0}^{\infty} a {r}^{n} - {\sum}_{n = 1}^{\infty} a {r}^{n} = a {r}^{0} = a$

If $\left\mid r \right\mid > 1$ then ${\sum}_{n = 0}^{\infty} a {r}^{n}$ will not converge.

One very useful infinite series that converges for any $x \in \mathbb{R}$ is

exp(x) = sum_(n=0)^oo (x^n)/(n!) = 1 + x/(1!) + (x^2)/(2!) + (x^3)/(3!) +...

The transcendental number $e$ is

exp(1) = 1+1/(1!)+1/(2!)+1/(3!)+... ~= 2.71828182844

In fact, you can show that $\exp \left(x\right) = {e}^{x}$