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?
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 $|z|=1$.