Let $\sum a_nz^n$ and $\sum b_nz^n$ power series with radiuses of convergences $R_1,R_2$ respectively. Suppose the radius of convergence of $(\sum a_n+b_n)z^n$ is $R$. Find an example in which $\infty>R>\min \{R_1,R_2\}$, given $R_1=R_2$.

What am I to do here? Previously I was required to show that $R=\{R_1,R_2\}$ (Given $R_1\ne R_2$, how does it matter?) and now that? Besides, $(\sum a_n+b_n)z^n=\sum a_nz^n+\sum b_nz^n$ and if $R$ is bigger the the minimum then one of the series must not converge, isn't that so? I would appreciate your help.


$$1+x+\frac{x^2}{2^2}+x^3 + \frac{x^4}{2^4} + x^5 + \frac{x^6}{2^6}+\cdots $$


$$1-x+\frac{x^2}{2^2}-x^3 + \frac{x^4}{2^4} - x^5 + \frac{x^6}{2^6}- \cdots$$

both have radius of convergence $1.$ But the sum of these series is

$$2+ \frac{x^2}{2} + \frac{x^4}{2^3}+ \frac{x^6}{2^5}+\cdots ,$$

which has radius of convergence $2.$

  • $\begingroup$ How is the radius of convergence computed? I have to find $a_n$ and then find the limit $a_n^{1\over n}$ $\endgroup$ – Meitar May 2 '15 at 17:15
  • $\begingroup$ Not really. In the first two series, we have convergence for $|x|<1,$ divergence for $|x|>1.$ There's only one number with that property, and its the radius of convergence. Same idea for the sum series. $\endgroup$ – zhw. May 2 '15 at 17:27

Consider the series for $\frac1{1-x}$ and the series for $\frac1{2-x}-\frac1{1-x}$. Their sum is the series for $\frac1{2-x}$.

  • $\begingroup$ Should I find $a_n$ and $z$ that will fit these series? $\endgroup$ – Meitar May 2 '15 at 16:41
  • 1
    $\begingroup$ @Meitar Hint: Geometric series. $\endgroup$ – Tim Raczkowski May 2 '15 at 16:47
  • $\begingroup$ If $a_n={1\over 1-n}$ and $b_n={1\over 2-n}-{1\over 1-n}$ then the radius of the sum is and the radius of $a_n,b_n$ is the same, isn't it? $\endgroup$ – Meitar May 2 '15 at 16:55
  • $\begingroup$ I understand now Tim, I guess $\endgroup$ – Meitar May 2 '15 at 17:00

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