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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}

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    What is your meaning?2012-05-27
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    Ok I see NOW! THANK YOU !!2012-05-27

1 Answers 1

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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|

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 $R_1=R_2$?2012-05-27
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    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 $R_1$?2012-05-27
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    The last series is $\sum_n c_nx^n$, whose radius of convergence is $R$ (so there is a typo).2012-05-27
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    In fact, identifying a sequence with the given power series, if $R$\{a_n\},\{b_n\}$ such that the radius of these series is $R$ but the radius of convergence of the sum is $R'$. – 2012-05-27
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    For a cheap example look at $f(z)=\sum_0 ^{\infty} z^n$ and let $g(z)=-f(z)$ then R.O.C of $f=$ R.O.C of $g=1$ But $f+g$ is identically zero and has infinite R.o.c .2014-10-31