Conditional Convergence in a Power Series

1. Jun 1, 2010

nonequilibrium

I was wondering if there's an example of a power series $$\sum_n^\infty c_n (z-a)^n$$ with radius of convergence R so that all z for which |z-a| = R there is purely conditional convergence? (no divergence but also no absolute convergence) Or perhaps a reason why that's impossible?

2. Jun 1, 2010

Mute

I'm not sure, but I might try taking a conditionally convergent series and put a (z-a)^n in the summand. For example, try

$$f(z) \equiv -\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(1-z)^n$$

Without the -(1-z)^n this is a conditionally convergent series. I picked -(1-z)^n to tack on because without the (-1)^(n+1) this series gives the logarithm when |1-z| < 1. For the logarithm, the series diverges beyond this radius. I would guess that the modified series I proposed might conditionally converge beyond that radius, instead of diverging, but I haven't checked for sure.

3. Jun 3, 2010

nonequilibrium

Hm, take z = 2, then you get the harmonic series and it diverges. Thanks for the try though.

Apparently one can proof that $$\sum \frac{z^n}{n}$$ converges conditionally FOR ALL |z| = 1 except for z = 1. That's awfully close, but sadly not enough :(

4. Jun 3, 2010

l'Hôpital

$$f(z) = \sum_{n=1}^{\infty} \frac{(-|z|)^n}{n}$$

Looks to me like it converges for all |z| < 1. Consider the consider you care about: |z| = 1 = R.

Of course, since |z| = |-z| the sign change won't affect convergence. And of course, if plugging |z|= 1, you get conditional convergence automatically!

Does that work for you?

Edit: Oops, misread. You wanted a power series. My bad. : (

5. Jun 3, 2010

nonequilibrium

Yeah a power series, sorry :( thanks for the effort though :)