What is the interval of convergence of #sum (x^n)/(n!) #?

2 Answers
Nov 4, 2015

#(-oo, oo)#

Explanation:

For any #x in RR#, choose #N in ZZ# such that #N > abs(x)#

#abs(sum_(n=0)^oo x^n/(n!)) = abs(sum_(n=0)^(N-1) x^n/(n!) + sum_(n=N)^oo x^n/(n!)) <= sum_(n=0)^(N-1) abs(x)^n/(n!) + sum_(n=N)^oo abs(x)^n/(n!)#

#< sum_(n=0)^(N-1) abs(x)^n/(n!) + abs(x)^N/(N!) sum_(n=0)^oo abs(x)^n/(N^n)#

The first term #sum_(n=0)^(N-1) abs(x)^n/(n!)# is a finite sum so converges.

The second term #abs(x)^N/(N!) sum_(n=0)^oo abs(x)^n/(N^n)# is the sum of a geometric series with positive common ratio #abs(x)/N < 1#, so converges.

We have shown that for any #x in (-oo,oo)#, #sum_(n=0)^oo abs(x)^n/(n!)# is bounded, that is that #sum_(n=0)^oo x^n/(n!)# is absolutely convergent. Hence it is also convergent.

Take the absolute values and apply the ratio test

#lim_(n->oo)abs(a_(n+1))/(abs(a_n))=lim_(n->oo) (abs(x)^(n+1)/((n+1)!))/(abs(x)^(n)/((n)!))=lim_(n->oo) absx/((n+1)!)=0#

The limit is less than 1, independent of the value of x. It follows that the series converges for all x.
That is, the interval of convergence is # −∞ < x < +∞#.

Actually the sum is equal to the exponential function

#Σ x^n/(n!)=e^x#