What is the interval of convergence of #sum ((-1)^n * x^(n+8) #?

1 Answer
Feb 25, 2016

Answer:

The radius of convergence #R# of a power series #sum_(n=0)^(oo) a_n x^n# is given by #R = 1/(lim "sup"_(n->oo)root(n)(abs(a_n)))#.

In that case, #R = 1#.

Explanation:

Let's write #sum_(n=0)^(oo) (-1)^n * x^(n+8)# in the form of #sum_(n=0)^(oo) a_n x^n#.

#sum_(n=0)^(oo) (-1)^n * x^(n+8) = sum_(m=0)^(oo) a_m x^m#,

where #a_m = 0 if m<8 and a_m = (-1)^(m-8) if m>=8#.

Now you can compute # R = 1/(lim "sup"_(m->oo)root(m)(abs(a_m)))#.

Yet, #abs(a_m)# can only take #0# or #1# as a value. But when #m# tends to infinity (so when #m# is much larger than #8#), #abs(a_m) = 1 = root(m)(abs(a_m)# so #lim "sup"_(m->oo)root(m)(abs(a_m)) = 1#.

Therefore, #R = 1/1 = 1#.

You now know that #sum_(n=0)^(oo) (-1)^n * x^(n+8)# converges for #x in ]-1, 1[#.

What about #x = -1# or #x = 1# ?

If #x = -1# :
#sum_(n=0)^(oo) (-1)^n * (-1)^(n+8) = sum_(n=0)^(oo) (-1)^(2n + 8) = sum_(n=0)^(oo) 1 = +oo#

If #x = 1# :
#sum_(n=0)^(oo) (-1)^n * (1)^(n+8) = sum_(n=0)^(oo) (-1)^n# is not defined.

So #sum_(n=0)^(oo) (-1)^n * x^(n+8)# converges for #x in ]-1, 1[#.