Can an infinite series have a sum?

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Answer 1 · 2015-07-31T18:29:06

Yes.

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

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 $exp(x) = e^x$