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Given two power series $$\sum_{n=0}^{\infty} a_nx^n, \sum_{n=0}^{\infty} b_nx^n$$ with convergent radius $R_{1}$ and $R_{2}$ respectively. Suppose $R_{1}<R_{2}$,now what about the convergent radius of $\sum_{n=0}^{\infty} (a_n+b_n)x^n$

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What is your meaning? –  Riemann May 27 '12 at 11:27
    
Ok I see NOW! THANK YOU !! –  Riemann May 27 '12 at 11:36

1 Answer 1

up vote 3 down vote accepted

For a power series $\sum_n c_nx^n$, the radius of convergence is $\sup\{R>0, \{c_nx^n\}\mbox{ is bounded if }|x|\leq R\}$.

Let $x$ such that $|x|<R_1$. Then $|(a_n+b_n)x^n|\leq |a_n||x|^n+|b_n||x^n|$ and the sequence $\{(a_n+b_n)x^n\}$ is bounded. Hence the radius of convergence of $\sum_n(a_n+b_n)x^n$ is $\geq R_1$. If we take $|x|\in (R_1,R_2)$, then $\{(a_n+b_n)x^n\}$ is not bounded, as a sum of a bounded sequence with an unbounded one. So the radius of convergence is at most $R_1$.

Conclusion: the radius of convergence of $\sum_n (a_n+b_n)x^n$ is $R_1$.

Now we look at the case $R_1=R_2$. Let $R>R_1$ and $\sum_nc_nx^n$ a series or radius of convergence $R$. By the previous case, $\sum_n(-a_n+c_n)x^n$ has a radius of convergence $R_1$ but $\sum_n(a_n+(-a_n+c_n))x^n$ has a radius of convergence $R$. So the sum of two series of radius of convergence $R_1$ can be of radius of convergence $R\geq R_1$.

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What about when $R_1=R_2$? –  Riemann May 27 '12 at 11:48
    
I think it is not clear in the case $R_1=R_2$, adn WHY $\sum_n(a_n+(-a_n+c_n))x^n$ has a radius of convergence $R_1$? –  Riemann May 27 '12 at 14:34
    
The last series is $\sum_n c_nx^n$, whose radius of convergence is $R$ (so there is a typo). –  Davide Giraudo May 27 '12 at 14:36
    
In fact, identifying a sequence with the given power series, if $R<R'$ are given number, we can find $\{a_n\},\{b_n\}$ such that the radius of these series is $R$ but the radius of convergence of the sum is $R'$. –  Davide Giraudo May 27 '12 at 14:53

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