I need to show that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+x^2}$ converges uniformly but not absolutely on $\mathbb{R}$.
First, I showed that the absolute value of the partial sums diverges for any value of $x \in \mathbb{R}$ by comparison to the harmonic series. But the series converges by the Alternating Series Test since $\frac{1}{n+x^2}$ is monotonically decreasing for any fixed $x \in \mathbb{R}$ and it vanishes as $n \rightarrow \infty$
Then I proceed to use the Cauchy Condition to show that the series has bounded partial sums. But this is where I think I messed up. This is what I have.
$$\left|\sum_{n=m+1}^{k}\frac{(-1)^{n+1}}{n+x^2}\right| \leq \left|\sum_{n=m+1}^{k}\frac{1}{n+x^2}\right| \leq \left|\sum_{n=m+1}^{k}\frac{1}{n}\right| \leq \epsilon$$
My confusion is that while I know that the series partial sum I've bounded my series by is divergent, but I also know that my series is always less than it. Am I doing anything right?
I fear that this is not quite correct.
So... I have the answer. We can't easily bound the series using Cauchy's Condition. The obvious choice of $\sum_{n=m+1}^{k}\frac{1}{n}$ doesn't have an upper bound. It diverges. Rather, there is a consequence of the Alternating Series Test that I missed, or ignored or forgot... the consequence is the following:
We know the series converges conditionally by the A.S.T. So We can call the sum A. If $s_n$ is the partial sum of our sereis, $$\left| A - s_{n}\right| \leq a_{n+1} $$ so we can make this more explicit:
$$\left|\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+x^2} - \sum_{n=1}^{k}\frac{(-1)^{n+1}}{n+x^2}\right| \leq \left|\frac{(-1)^{n+1}}{n+x^{2}}\right| \leq \frac{1}{n+x^2} \leq \frac{1}{n} \leq \epsilon$$ for any $x \in \mathbb{R}$.
RRL's answer above using Dirichlet's Test also works and arrives at the same conclusion. This answer just completes where I started off instead of starting over.