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}
Radius of convergence of Power Series!
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0Ok I see NOW! THANK YOU !! – 2012-05-27
1 Answers
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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0For 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