If $\{a_n\}$ is a sequence of real numbers for which $\sum a_{2n}$ and $\sum a_{2n+1}$ both converge, must $\sum a_n x^n$ converge uniformly on $[-1,1]$?
I've convinced myself that the full series $\sum a_n$ must converge because it is Cauchy, something like: $$\left | \sum a_n \right | \leq \left | \sum a_{2k} \right | + \left | \sum a_{2k+1} \right |,$$ and since the even and odd series converge, the sums on the right hand side can be made arbitrarily small for sufficiently large $N$.
But this means the $\limsup |a_n|^{1/n} \leq 1$, which says that $\limsup |a_n x^n|^{1/n} \leq |x|$, so the power series $\sum a_n x^n$ must converge at least in $(-1,1]$, although the radius of convergence could be larger (take $a_n = 2^{-n}$).
Since the radius of convergence of $\sum a_n x^n$ is at least 1, we know that the series $\sum a_n x^n$ must converge uniformly in $(-1 + \epsilon, 1-\epsilon)$, for any $\epsilon > 0$, but I am stuck showing uniform convergence in all of $[-1,1]$.
If I could say that $\left | \sum a_{2n}x^{2n} \right | \leq \left | \sum a_{2n} \right |$ for $x\in[-1,1]$, I could then say
\begin{align*} \left | \sum a_n x^n \right | &\leq \left | \sum a_{2k} x^{2k} \right | + \left | \sum a_{2k+1} x^{2k+1} \right |\\ &\leq \left | \sum a_{2k} \right | + \left | \sum a_{2k +1} \right |,\end{align*}
and get uniform convergence in [-1,1] by the Cauchy condition, but I cannot convince myself that the needed inequality is true.
The best I can do with the Weirstauss M-Test is $|a_n x^n| \leq |a_n|$ when $x \in [-1,1]$, but since we do not know that $\sum a_n$ converges absolutely, this is not very helpful.
But this does mean to look for a counter-example, we need to look for a sequence $\{a_n\}$ for which $\sum a_{2n}$ and $\sum a_{2n+1}$ converge, and for which $\sum a_n$ converges conditionally. I've played around with variations on alternating harmonic series, and adaptations of $\sum \frac{x^n}{2^n}$, which does not converge uniformly in all of $(-2,2)$, with no luck.
Gentle hints appreciated.