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 $|z-a|<R$ 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?

  • $\begingroup$ 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. $\endgroup$ – 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 $|z|=1$.

  • $\begingroup$ 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? $\endgroup$ – Mathias May 15 '20 at 12:45
  • 1
    $\begingroup$ Yes, it is just the contrapositive statement. @Mathias $\endgroup$ – Kavi Rama Murthy May 15 '20 at 12:47
  • $\begingroup$ Thank you Kavi for clarifying! $\endgroup$ – Mathias May 15 '20 at 12:50

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