What is the interval of convergence of #sum_1^oo [(-x)^(2n+1)]/[(2n+1)!]#?

1 Answer
Jan 5, 2017

The interval of convergence is # x in ] -oo,+oo [ #

Explanation:

We use the ratio test

If the #lim_(n->oo)∣a_(n+1)/a_n∣=L#

If #L<1#, then #sum_1 ^oa_n# converges

If #L>1#, then #sum_1 ^oa_n# diverges

If #L=1#, the test is inconclusive

#(∣a_(n+1)/a_n∣) =(∣((-x)^(2(n+1)+1)/((2(n+1)+1)!))/((-x)^(2n+1)/((2n+1)!))∣)#

Now, we calculate the limits

#lim_(n->oo)(∣((-x)^(2(n+1)+1)/((2(n+1)+1)!))/((-x)^(2n+1)/((2n+1)!))∣)#

#=lim_(n->oo)(∣((-x)^(2n+3)/((2n+3)!))/((-x)^(2n+1)/((2n+1)!))∣)#

#=lim_(n->oo)(∣((-x)^(2n+3)/((-x)^(2n+1))*(((2n+1)!)/((2n+3)!))∣)#

#=lim_(n->oo)(∣x^2*(1/((2n+3)(2n+2))∣)#

#=lim_(n->oo)(∣x^2*(1/((2n+3)2(n+1))∣)#

#=∣(x^2/2)∣*lim_(n->oo)(∣1/((2n+3)(n+1))∣)#

#=∣(x^2/2)∣*0#

#=0#

Therefore,

#L<1#, for every x,

So #sum_1 ^oox^(2n+1)/((2n+1)!)# converges for all x

The interval is # -oo < x <+oo #