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Problem: Suppose the power series $ \sum^\infty_{n=0}{ a_n x^n } $ converges at radius of convergence $ R_a $ and the power series $ \sum^\infty_{n=0}{ b_n x^n } $ converges at radius of convergence $ R_b $.

What can be said about the radius of convergence of the power series $ \sum^\infty_{n=0}{ (a_n+b_n) x^n } ~~ $ ?
Hint: Handle the case $ R_a = R_b $ and the case $ R_a \neq R_b $ separately.

Attempt- I didn't really know what to do but I still tried:
If $ R_a \neq R_b $ the sum $ \sum^\infty_{n=0}{ (a_n+b_n) x^n } $ converges iff $ \sum^\infty_{n=0}{ a_n x^n } $ and $ \sum^\infty_{n=0}{ b_n x^n } $ converge, hence $ R_{ab} = min\{ R_a , R_b \} $.
Let's now look at a case where we'd have $ R_a = R_b $; Define $ S^{a}_n = \sum^{n}_{k=1}(-1)^k x^k $ , $ S^{b}_n = \sum^{n}_{k=1}(-1)^{k+1} x^k $ note that $ R_a = R_b =1 $ . But $ S^{ab}_n = \sum^{n}_{k=1}( (-1)^k + (-1)^{k+1} ) x^k = 0 $ hence $ R_{ab} = +\infty $ .

Question: I feel very lost at this problem, can you please help me as to what explicitly I should do and what to show? how would you approach this problem?

hazelnut_116
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    I think you ought to write your argument for the first case out in a little more detail; that may point you in the right direction for the equality case. Why does the sum converge only if $|x|\leq \min{R_a,R_b}$? – saulspatz May 17 '21 at 19:10
  • Can you write $\sum (a_n +b_n)x^n$ as $\sum a_n x^n +\sum b_n x^n $? – Vishu May 17 '21 at 19:20
  • @Tavish Yes, If both power series converge at overlapping radii of convergence. – hazelnut_116 May 17 '21 at 19:36
  • @saulspatz The $ \sum (a_n +b_n)x^n $ converges for a radii smaller or equal to $ min { R_a, R_b } $ since both $\sum a_n x^n $ ,$\sum b_n x^n$ converge for that radii – hazelnut_116 May 17 '21 at 19:39
  • That's the if part, not the only if part. – saulspatz May 17 '21 at 19:58

2 Answers2

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The point I was trying to make in a comment is this. Suppose $R_b>R_a$. Then certainly $\sum (a_n+b_n)x^n$ converges in $|x|< R_a$. Now suppose that $R_a<|x|<R_b$. Then $\sum (a_n+b_n)x^n$ can be written as the sum of a convergent series and a divergent one, and hence diverges. Therefore, the radius of convergence can't be any bigger than $R_a$ and we see that $R_{ab}=\min\{R_a,R_b\}$.

When $R_a=R_b$ we still have $R_{ab}\geq R_a$, but we can't produce a point at which we know the sum diverges, as in the above argument. Outside $|x|\leq R_a$, we have the sum of two divergent series, and we can't say anything definite. If fact, let $R>R_a$ and let $\sum c_nx^n$ be a power series with radius of convergence $R$. Now define $b_n = c_n-a_n$. By the first part, $R_b=\min\{A_a,R\}=R_a$, but $R_{ab}= R$. The only thing we can say is that $R_{ab}\geq R_a$.

saulspatz
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Generally you can only say that if $|x| < \min \{R_a, R_b\}$, then $\sum_n (a_n + b_n)x^n$ must converge. However, there are examples where it converges on a wider interval (example $a_n = -b_n$).

Gábor Pálovics
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