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$R_1$ Is the Radius of convergence of $\sum_{k=1}^{\infty}a_nx^n$.
$R_2$ Is the Radius of convergence of $\sum_{k=1}^{\infty}b_nx^n$.
What can we say of the Radius of convergence of $\sum_{k=1}^{\infty}(a_n+b_n)x^n$,we'll call it $R$?


If $R_1 \neq R_2$ I know the answer is $R=M=\min(R_1,R_2)$ Because it is obvious that for all $x \in (-M,M)$ the sum converges, and suppose $R>M$ (And without loss of generality $R_2>R_1$), than take new number $L>0$ Such that $L<R$ and $R_1<L<R_2$, than since $L<R$ we get $\sum_{k=1}^{\infty}(a_n+b_n)L^n$ converges, and that's a contradiction because $\sum_{k=1}^{\infty}a_nx^n$ diverges (because L is bigger than $R_1$) but $\sum_{k=1}^{\infty}b_nx^n$ converges (because L is bigger than $R_2$). so $R=M$.

But what if $R_1=R_2$? we can take for example $a_n=1/n$, $b_n$=$-1/n$ and then we'll get $R=\infty$. We can take $a_n=b_n=1/n$ and than simply $R=R_1=R_2$. Is there anything we can say about this case? is it always $R=R_1=R_2$ or $R=\infty$ in this case?

thanks!

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Yes: you can say that $R\geqslant R_1=R_2$. But that's all you can say.

Take for instance, the series $\displaystyle\sum_{n=0}^\infty\left(1+2^{-n}\right)z^n$ and the series $\displaystyle\sum_{n=0}^\infty-z^n$. The radius of convergence of both of them is $1$, but the radius of convergence of their sum is $2$.

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  • $\begingroup$ Thanks! Is there an example where R>R1=R2 (and not equale) but R is not infinity? $\endgroup$ – Omer Dec 10 '18 at 14:28
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    $\begingroup$ I've added one such example to my answer. $\endgroup$ – José Carlos Santos Dec 10 '18 at 14:29
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    $\begingroup$ No. The radius of convergence of $\sum_{n=0}^\infty\frac{z^n}{2^n}$ is $2$. $\endgroup$ – José Carlos Santos Dec 10 '18 at 14:47
  • $\begingroup$ Yes, you are right, I just realized that and deleted my comment a second before you answered. Thanks, you helped me a lot! $\endgroup$ – Omer Dec 10 '18 at 14:48

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