Radius of convergence and sum of series $3nz^{n-1}$

Question

I'm learning complex series and I have a doubt.

I have to find the radius of convergence and the sum of $$\sum_{2}^{\infty} 3nz^{n-1}$$

For $z=0$ I believe the series converges and the sum is $0$.

For $z\neq0$ I've tried to apply the ratio test but with no success. This is what I have done. $$\lim_{n \to \infty} \sqrt[n]{|3n\frac{z^n}{z}|}\Rightarrow \lim_{n \to \infty} |z|\sqrt[n]{\frac{3n}{z}} $$

Getting here I've realized that I've must have done something wrong.

Can someone point me in the right direction?

Ideas for finding the sum can be found in this post and the posts in its "related" section. — David Mitra, Jun 11 '13 at 18:10

score 4 · Accepted Answer · answered Jun 11 '13 at 18:29

4

Try the quotient test (many times more gentle than the $\,n$-th root one):

$$\left|\;\frac{3(n+1)z^n}{3nz^{n-1}}\;\right|=\frac{n+1}n|z|\xrightarrow[n\to\infty]{}|z|<1$$

answered Jun 11 '13 at 18:29

DonAntonio

211,718
17
136
287

Thanks. Forgot to use that test :( And if $|z|<1$ what would be the radius of convergence? Before making any tests can I assume if $z=0$ that the series is convergent? – Favolas Jun 11 '13 at 18:39

score 0 · Answer 2 · answered Jun 11 '13 at 18:08

0

It will be nth root of $(3n)$

So the radius of convergence will be $1$ as $n^{1/n}\to 1$ and $3^{1/n}\to 1$as $n\to \infty$.

answered Jun 11 '13 at 18:08

Abhra Abir Kundu

8,150

Thanks. But where did the z's go? – Favolas Jun 11 '13 at 18:19
You can denote $y_{i}=x_{i+1}$ ana then apply root test to ${y_i}$ – Abhra Abir Kundu Jun 11 '13 at 18:21

Radius of convergence and sum of series $3nz^{n-1}$

2 Answers2