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I have two series

$\displaystyle\sum_{n=1}^{\infty} a_n x^{n}$

$\displaystyle\sum_{n=1}^{\infty} b_n x^{n}$

with radius of convergence $2$ and $3$ respectively.

How can I find the radius of convergence for the following?

$\displaystyle\sum_{n=1}^{\infty} (a_n +b_n)x^{n}$

Can anyone help me?

rschwieb
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Dimitri
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2 Answers2

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You can show that it is the minimum of the 2 convergence radius, if the are different. This one is easy as for $x\leq \min\{\rho_1,\rho_2\}$ the convergence of the sums is absolute. So it is just simple arithmetic $$\sum_{n=1}^\infty (a_k +b_k ) x^k = \sum_{n=1}^\infty a_k x^k + \sum_{n=1}^\infty b_k x^k$$

When they are the same, you only can say that it is greater equal than the convergence radius. Taking for example $a_k=-1$ and $b_k=1$ the convergence radius of $$ \sum_{k=1}^\infty (a_k+b_k) x^k $$ is infinity.

To see that if the radius are different we really only have the minimum and not more as the convergence radius we make this here:

Let us say the convergence radius of $\sum_{k=1}^\infty a_k x^k $ is $\rho_1$ and the one of $\sum_{k=1}^\infty b_k x^k$ is $\rho_2$ we say now $\rho_1 < \rho_2$,

If the convergence radius would be greater than $\rho_1$ than the following equation must be true: $$\sum_{k=1}^\infty (a_k +b_k) x^k -\sum_{k=1}^\infty b_k x^k = \sum_{k=1}^\infty a_k x^k. $$ As on the left hand side we have the difference of two convergent series the right hand side must be convergent too, but it is only convergent if $x\leq \rho_1$.

Dominic Michaelis
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  • so, is not the minimum of the 2 radius in general, as i though, but i need an idea for that particular case – Dimitri Feb 20 '13 at 20:34
  • For example, for $2 < |x| < 3$, we have $|a_n x^n| \to \infty$ and $|b_n x^n| \to 0$, so $|(a_n+b_n)x^n| \to \infty$. – GEdgar Feb 20 '13 at 20:34
  • @Dimitri it is the minimum if they are different else I don't know (at least they are 2 cases, radius infty and or of the old one) – Dominic Michaelis Feb 20 '13 at 20:37
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Since

$\displaystyle\sum_{n=1}^{\infty} (a_n +b_n)x^{n}=\displaystyle\sum_{n=1}^{\infty} a_n x^{n}+\displaystyle\sum_{n=1}^{\infty} b_n x^{n} $ it is clear that the left hand side converges for values of $x$ exactly when both of the series on the RHS converge.

Since this means that the LHS will converge on the intersection of the discs of convergence on the RHS, you know the radius of convergence for the LHS is the same as the smaller radius on the RHS.

I only have confidence in this answer for absolutely convergent series. Perhaps its necessary after all to resort to a test for convergence, rather than relying on this naive arithmetic.

rschwieb
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    Actually I haven't had this topic in a decade now, so I may have forgotten some quirks. It seems like this naive logic is sufficient, but I appreciate any illuminating remarks if there is some flaw. It will definitely converge within 2, and it will definitely diverge between 2 and 3, so the radius must be 2. – rschwieb Feb 20 '13 at 20:43
  • I think that the LHS can converge, without need that the series in RHS converge (for example if $a_n=-b_n$ the LHS have infinite radius, but both series of RHS don´t even need to converge – Dimitri Feb 20 '13 at 20:46
  • @rschwieb don't forget it doesn't have to be absolute convergent, if you change sum order you can change the convergence behaviour look at my example – Dominic Michaelis Feb 20 '13 at 20:47
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    @Dimitri If $a_n=-b_n$, you will definitely not have radii of convergence 2 and 3. – rschwieb Feb 20 '13 at 20:49
  • @DominicMichaelis But that example does not have the right radii of convergence... can it be adjusted so that it does? I think you're right I have to double check what goes on with conditionally convergent series.. – rschwieb Feb 20 '13 at 20:50
  • @rschwieb no it can't as we know that $a_n x^n$ will be no $0$- sequence for $2<|x|<3$ by reordering we have $$ \sum_{n=1}^\infty (a_n+b_n) x^n - \sum_{n=1}^\infty b_n x^n = \sum_{n=1}^\infty a_n x^n$$. As we are saying $\sum_{n=1}^\infty b_n x^n $ does converge and $\sum_{n=1}^\infty (a_n+b_n)x^n $ we got on the left hand side the difference of two convergent things, but on the right hand side a divergent one that can'T be – Dominic Michaelis Feb 20 '13 at 20:55
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    @DominicMichaelis For sure I see now why the simple "addition proof" breaks down for conditionally convergent cases :) Thanks for refreshing my series knowledge. – rschwieb Feb 20 '13 at 20:59