Question

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

Answer 1

`(-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`.