# Radius of convergence of two power series and the sum of the power serieses

$$R_1$$ 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 $$L and $$R_1, than since $$L we get $$\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!

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 $$2$$.
• No. The radius of convergence of $\sum_{n=0}^\infty\frac{z^n}{2^n}$ is $2$. – José Carlos Santos Dec 10 '18 at 14:47