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$$\sum_{n=1}^{\infty} \frac{n^3}{3^n} z^n$$ As a power series in $z$, it has a radius of convergence, which is what I believe you are looking for.

Regroup it as $\displaystyle n^3 \left( \dfrac{z}{3} \right)^n$.

Use the ratio test: In a power series $\displaystyle \sum_{n=1}^{\infty} a(n) x^n$, if $\displaystyle \lim_{n \rightarrow \infty} \left \lvert \dfrac{a(n+1)}{a(n)} \right \rvert = c$, then the series converges for $\displaystyle \lvert x \rvert < \frac1c$.

Taking $x = z/3$, you see that $a(n) = n^3$.

$$\lim_{n \rightarrow \infty} \left \lvert \dfrac{a(n+1)}{a(n)} \right \rvert = \lim_{n \rightarrow \infty} \left( 1 + \frac1n\right)^3 = 1$$

So convergence occurs for $|z| < 3$.

The question of convergence remains open for the endpoints, $z = \pm 3$ of the convergence interval. This is settled by realizing that at both those endpoints, the magnitudes of the terms are all greater than those of the corresponding terms of $\displaystyle \sum_{n=1}^{\infty} \left( \pm x \right)^n$ and that both of these are non-convergent. So by the comparison test, convergence fails at the endpoints, and is strictly confined to the interior of the interval $|z| < 3$.

Is it right to my procedure?

Is there another easier way to do it?

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Kindly refer here ( on how to typeset (i.e. how to write equations etc) on this website. Typesetting makes it easier to read. – user17762 May 22 '12 at 4:48
Are we just pre-checking your series homework assignment for you this evening? In the past hour we've had this question, this, this, and this. – Arturo Magidin May 22 '12 at 6:09
up vote 2 down vote accepted

Your argument for $\lvert z \rvert < 3$ looks fine. On the boundary points, i.e. $z = \pm 3$, I am not exactly sure what your argument is. For $z = +3$, the series becomes $$\sum_{n=1}^{\infty} n^3$$ and at $z = -3$, the series becomes $$\sum_{n=1}^{\infty} (-1)^n n^3$$ Remember that if a series of the form $\displaystyle \sum_{n=1}^{\infty} a_n$ converges, then $\displaystyle \lim_{n \rightarrow \infty} a_n = 0$ (or) to put it the other way if $\displaystyle \lim_{n \rightarrow \infty} a_n \neq 0$, then the series $\displaystyle \sum_{n=1}^{\infty} a_n$ doesn't converge.

Make use of this to obtain your conclusion.

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