Question

Can an infinite series have a sum?

Answer 1

Yes.

For example `1+1/2+1/4+1/8+1/16+... = sum_(n=0)^oo2^-n=2`

Explanation:

In general, if `r in (-1, 1)` and `a != 0`, then:

`sum_(n=0)^oo ar^n = a/(1-r)`

To see this notice that:

`(1-r)sum_(n=0)^oo ar^n = sum_(n=0)^oo ar^n - r sum_(n=0)^oo ar^n`

`= sum_(n=0)^oo ar^n - sum_(n=1)^oo ar^n = ar^0 = a`

If `abs(r) > 1` then `sum_(n=0)^oo ar^n` will not converge.

One very useful infinite series that converges for any `x in RR` 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(x) = e^x`