Can an infinite series have a sum?

1 Answer
Jul 31, 2015

Answer:

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#