# domain of convergence of series $\sum_{n=0}^{\infty}n^2(2z−1)^n$

I need to find domain of convergence of series $$\sum_{n=0}^{\infty}n^2(2z−1)^n$$.
What I did is I take $$2^n$$ out like $$\sum_{n=0}^{\infty}n^2 2^n(z−1/2)^n$$.
Let $$z_1=z-1/2$$ make the series like $$\sum_{n=0}^{\infty}n^2 2^n(z_1)^n$$.
Then I let $$c_n=n^2 2^n$$ and found $$R=1/2$$. So domain of convergence for $$z_1$$ is $$D(0,1/2)$$. If I want to translate it to $$z-1/2$$ is it like shifting the ball to the right by 1/2 on the real axis like how we did it in $$\mathbb{R^2}$$ and $$D(1/2,1/2)$$?

Is my reasoning correct? Thanks for your help!!

It is correct but you could use the ratio test from the begining. Put $$u_n=n^2(2z-1)^n$$ then

$$\lim_{n\to+\infty}\frac{|u_{n+1}|}{|u_n|}=|2z-1|$$

So $$|2z-1|<1\iff |z-\frac 12|<\frac 12\implies \sum u_n \text{ converges}$$

and

$$|2z-1|>1\iff |z-\frac 12|>\frac 12\implies \sum u_n \text{ diverges }$$

we conclude that $$R=\frac 12$$.

• How did you conclude that |2z-1|<1 and conclude that it converges?. I didn't understand that part. Apr 3 at 22:37
• @Mrnobody I did not conclude that. the ratio test says that if the limit is $<1$, the series converges. Apr 3 at 22:38
• now I understand thank you very much!! Apr 3 at 22:39

Using the $$n$$-th root test one gets convergence fo al $$z$$ such that $$\lim_n\sqrt[n]{n^2}|2z-1|=2|z-1/2|$$, that is for all $$z$$ inside the circle centered at 1/2 and radius $$1/2$$

• thank you very much! Apr 3 at 22:39