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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!!

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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$.

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

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  • $\begingroup$ thank you very much! $\endgroup$
    – Mrnobody
    Apr 3 at 22:39

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