Series expansions with alternating series from endpoints: conjecture

A student of mine made the following two conjectures after working through examples of checking endpoints of series expansions, and although I feel that the conjectures are both false, I have not come up with counterexamples. I would appreciate someone doing so, or if there are none, telling me how to prove either or both of the conjectures.

Conj. $1$: "If a series expansion $E(x)$ has an interval of convergence with endpoints $a,b$, so that $E(a), E(b)$ are alternating series, then at most one of them converges absolutely.”

Conj. $2$: same, without "absolutely". Thanks.

• Must $E(x)$ be a power series? Or can it be any kind of series? Jan 24, 2016 at 6:20
• Is $\sum (-1)^n \frac{x^{2n}}{n^2}$ an alternating series? Jan 24, 2016 at 8:02
• David: any. André Nicolas: yes, that would do it. But see my following comment. Jan 24, 2016 at 11:31
• @nomadreid does my question answer your concerns? If so, you should accept it Jan 27, 2016 at 15:14
• @Stella Biderman Yes, thanks, it answered my concerns. I would like to formally accept it, but I don't see anything on the site that says "accept". ??? Jan 28, 2016 at 16:36

I'm going to assume that this is about power series, likely arising out of Taylor Polynomials.

The conjecture that it cannot be that both endpoints converge, but only one converges absolutely is correct. To see this, first note that the interval of convergence is symmetric around the center of the power series. This follows from the fact that the ratio test gives an expression of the form $|x-c|<r$ where $c$ is the center and $r$ is the radius of convergence.

Consider $\sum a_i(x-c)^i$, which converges on $(c-r,c+r)$ by the ratio test and possibly at the end points. To see if it converges at the end points, we plug them in to get $\sum a_i((c-r)-c)^i=\sum a_i(-r)^i$ and $\sum a_i((c+r)-c)^i=\sum a_i(r)^i$. When you take the absolute value of the terms of these two sums, you get the same thing, $a_ir^i$, and so any behavior defendant solely on the absolute value of the terms will be the same at both end points, which includes absolute convergence.

Old Answer Below This Line Both conjectures are false. There is an interesting connection here, which is that one of them is guaranteed to alternate but it's possible for them to have almost any behavior. Consider the following examples:

$$f(x)=\sum_{I=0}^\infty \frac{x^n}{n^2 2^n}$$

$f(x)$ converges for $-2<x<2$ by the ratio test, but since $\sum\frac{1}{n^2}$ converges, it converges absolutely at both end points.

We can get almost any end behavior we like, by using $n^{1/2}$ we get divergence at both endpoints and $n$ gives us convergence at exactly one, but not absolute convergence. The only thing we can't do is get one to converge absolutely and the other to diverge or to not converge absolutely, as they are guaranteed to be the same up to sign as noted above.

EDIT: On an unrelated note, if your student is familiar with complex numbers, tell them to experiment with plugging complex numbers into the power series. It turns out that there is a very simple way to determine the radius of convergence of a power series by looking at the function it represents, and not even the power series at all.

• Oops, sorry, I looked at my notes again and found that the student really conjectured that one could not get a series where both endpoints converged, but only one converged absolutely. Your example does not suffice to disprove this conjecture, since both endpoints are absolutely convergent. Of course, in your example, centered at 0, |E(a)| = |E(b)| but this is not going to be true in general, so I need a slightly different example for an interval of convergence with one endpoint converging absolutely and the other endpoint converging conditionally. Jan 24, 2016 at 12:34
• @nomadreid I edited to answer this question. It turns out that it is true in general. Jan 24, 2016 at 15:34