# Find the convergence radius of a power series and determine whether or not is convergent

Consider the power series

$$S = \sum_{n=0}^\infty n(1-2^{-n})z^n$$ Then I have to determine the convergence radius $$R$$ and to argue for whether or not the power series is convergent for $$|z|=R$$. To find the convergence radius $$R$$ I have used the ratio test find that $$\frac{|a_n|}{|a_{n+1}|} = \frac{|n(1-2^{-n})|}{|(n+1)(1-2^{-n+1})|}$$ where both the numerator and denominator tends to infinity when n tends to infinity. Thus we can use L'Hopitals rule: $$\frac{|n(1-2^{-n})|}{|(n+1)(1-2^{-n+1})|} \sim \frac{|1-2^{-n}+ln(2)2^{-n}n|}{|1-2^{-n+1}+ln(2)2^{-n+1}(n+1)|} \rightarrow_{n \rightarrow \infty} \frac{1}{1} = 1$$ which means that the power series has convergence radius $$R = 1$$.

But now my books says that the power series converges absolutely (it doesn't say anything, at least what I can found about converges only) if $$|z-a| but I am not sure what a is? I know that a power series has the form $$\sum_{n=0}^\infty a_n(z-a)^n$$ but does this mean that I have to rewrite $$S$$ in order to find a? I am little bit confused. On the internet, I found that $$S$$ converges when $$|z| < 1$$ thus it will not convergence as $$|z| = R = 1$$? Is this correct?

• As I know you should include z at the ratio test. And then the equation would be |z|<1. Then you need to substitute when z=1 into the series and see if the series convergence, then you substitute z=-1, into the series and see if the series converges. – Kasiopea May 15 '20 at 12:22

Your computation of $$R$$ is correct. But the value of $$R$$ does not tell what happens when $$|z|=R$$. In this case $$|n(1-2^{-n})z^{n}|=n(1-2^{-n})\to \infty$$ and this implies that the series does not converge when $$|z|=1$$.

• Do I have to use contraposition to conclude this? Because I know if a series $\sum_{k=1} a_k$ is convergent then the sequence $\{a_k\}_{k=1}^\infty$ converges to zero. Thus if $a_k$ does not converge to zero, the series does not converge or how do I conclude this? – Mathias May 15 '20 at 12:45
• Yes, it is just the contrapositive statement. @Mathias – Kavi Rama Murthy May 15 '20 at 12:47
• Thank you Kavi for clarifying! – Mathias May 15 '20 at 12:50