An unproven proposition in my book states that if the series of $a_{n}z^n$ has radius of convergence $R_1$ and the series of $b_{n}z^n$ has radius $R_2$. Then the radius of convergence of $(a_{n}+b_{n})z^n$ is either $min(R_{1},R_{2})$ or (if $R_1=R_2$) greater or equal to $R_1$
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3Good to know ${{}}$ – Apr 21 '16 at 18:25
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@AhmedHussein This looks like a premature assessment to me. You seem to have missed the word 'unproven', so this is far from being known. – Thomas Apr 21 '16 at 18:39
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@Thomas I was joking about the absence of a question. – Apr 21 '16 at 18:48
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@AhmedHussein Yes, I'm aware of this. – Thomas Apr 21 '16 at 18:50
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We have $a_nz^n+b_nz^n=(a_n+b_n)z^n$. Let $z \in \mathbb{C}$ such as $|z| < \min(R_1,R_2)$. $\sum a_n z^n$ and $\sum b_n z^n$ are absolutely convergent, so $\sum (a_n+b_n) z^n$ is convergent and we have $\sum (a_n+b_n) z^n=\sum a_n z^n + \sum b_n z^n$.
Since $\sum (a_n+b_n) z^n$ converges $\forall |z| < \min(R_1,R_2)$, we have that the radius of convergence of the sum is $\geq \min(R_1,R_2)$.
Bérénice
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