$\sum_{n=0}^\infty \left(\frac{(z+1-i)^n}{3^nn^2}\right)$ and the ratio test

Question

I'm given the example of $$\sum_{n=0}^\infty \left(\frac{(z+1-i)^n}{3^nn^2}\right)$$

The book tells me that using ratio test we get

$$p = \lim_{n\to\infty} \left|\frac{z+1-i}{3} * \frac{n^2}{(n+1)^2}\right| = |\frac{z+1-i}{3}|$$

How the heck did the book get that answer for the ratio test on that example? I got a completely different answer. I don't understand how they were able to bring the $n^2$ to the numerator or any part of the $n^2/((n+1)^2)$

Answer 1

If what you're trying to get is the convergence radius, then Cauchy-Hadamard formula works really nice here:

$$\lim\sup_{n\to\infty}\sqrt[n]{\frac1{3^nn^2}}=\lim\sup_{n\to\infty}\frac1{3\sqrt[n]{n^2}}=\frac13\implies R=3$$

and we get (absolute if you want) convergence in the disk $\;|z+1-i|<3\;$ .

Answer 2

How the heck did the book get that answer for the ratio test on that example?

Do you understand what is ratio test?

The usual form of the test makes use of the limit

$$ {\displaystyle L=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|.} $$ Now what are the $a_n$'s in your series?

I don't understand how they were able to bring the $n^2$ to the numerator or any part of the $n^2/((n+1)^2)$.

For a simpler version of your question, do you know how to use the ratio test to show that $$ \sum_{n=1}^\infty\frac{1}{3^nn^2}<\infty? $$