If I have a power series:($z$ is complex here)

$\displaystyle \sum_{n=0}^{\infty}z^n$ valid for |$z|<1$ and another

$\displaystyle \sum_{n=0}^{\infty}({\frac{z}{2}})^n$ valid for |$z|<2$

I understand they are both valid for |$z|<1$ but why is their sum

$\displaystyle \sum_{n=0}^{\infty}z^n+$$\displaystyle \sum_{n=0}^{\infty}({\frac{z}{2}})^n$ valid for |$z|<1$?

  • $\begingroup$ I think you are asking why the sum doesn't converge for any $|z|>1$. But if $|z|>1$ then the terms don't go to zero, so the series can't possibly converge. $\endgroup$ – Gerry Myerson May 4 '15 at 13:06

It is a general fact that the sum of two converging series is again convergent. This is a direct consequence of the fact that the sum of two converging sequences is convergent.

Conversely, the sum of divergent series and a converging series is divergent. Thus, if the radii of convergence are different, than for every $z$ with $|z|$ between the two radii the series is divergent. Showing, that the radius of convergence is at most the smaller of the two.

However, if the series both have the same radius of convergence, then there sum might have a larger radius of convergence. For an example consider $\sum_{n} z^n$ and $\sum_{n}(-1 +1/n!) z^n$. They both have radius of convergence $1$ but the sum converges everwhere.

For power series, one could also analyse the formulas for the radius of convergence, however, this would be rather more complicated. But to illustrate:

$$\sqrt[n]{|a_n+b_n|} \le \sqrt[n]{2 \max{(|a_n|,|b_n|)}} = \sqrt[n]{2} \sqrt[n]{\max{(|a_n|,|b_n|)}} = $$

$$\sqrt[n]{2} \max ( \sqrt[n]{|a_n|}, \sqrt[n]{|b_n|)} ) $$ Since $\lim_{n\to \infty} \sqrt[n]{2} =1$ it follows that $$\limsup\sqrt[n]{|a_n+b_n|} \le \max ( \limsup \sqrt[n]{|a_n|}, \limsup \sqrt[n]{|b_n|)} ) $$

The reciprocal of the left is the radius of convergence of the sum, which is thus larger than the smaller of the two radii (the reciprocal of the right).

  • $\begingroup$ I get that the sum of two convergent power series is convergent. I just don't understand why the radius of convergence of the sum is the region in which they are both individually valid. $\endgroup$ – Snickett May 4 '15 at 13:03
  • $\begingroup$ @Snickett this is also not true in general. It is only true when the radii of convergence are different, as in your case. I will add a sentence to clarify this. $\endgroup$ – quid May 4 '15 at 13:07

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