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

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Answer 1 · 2015-11-19T19:26:54

$(-3, 3)$

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 $1$ .

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