What is the interval of convergence of #sum (n^3)(x^(3n))/(3^(3n)) #?

1 Answer
Nov 19, 2015

#(-3, 3)#

Explanation:

If #x in (-3, 3)#, then #abs(x^3/3^3) < 1#, so #1 - abs(x^3/3^3) > 0#

Let #delta = 1 - abs(x^3/3^3) > 0#

Let #N = ceil(7/delta)#

Suppose #n >= 1#:

#(n+1)^3/n^3 = (n^3+3n^2+3n+1)/n^3#

#= 1+3/n+3/n^2+1/n^3 <= 1+7/n#

So if #n >= N#, then #(n+1)^3/n^3 <= 1+7/ceil(7/delta) <= 1+delta#

So #(n+1)^3/n^3 abs(x^3/3^3) <= (1+delta)(1-delta) = 1-delta^2 < 1#

Then from #N# onwards, the series converges faster than a geometric series with common ratio #1-delta^2#.

On the other hand, if #abs(x) >= 3#, then the series diverges faster than a geometric series with common ratio #1#.