$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{x^2+\sqrt{k}}$
Why doesn't it converge absolutely? I know it converges pointwise by alternating series test.
For uniform convergence:
I tried approximating $|R_n(x)|$ and got :
$|R_n(x)| \leq \frac{1}{x^2+\sqrt{n+1}}$ which goes to $0$, so $\|R_n\|_\infty \longrightarrow 0$. Therefore the series uniformly converges. Is this right? And what about absolute convergence please?
Non-absolute convergence: For all sufficiently large $k$ we have $x^2+\sqrt k\;<2\sqrt k\;....$ So for all but finitely many $k$ we have$$\left|\frac {(-1)^{k+1}}{x^2+\sqrt k\;}\right|>\frac {1}{2\sqrt k}\;.$$ And $\sum_k\frac {1}{2\sqrt k}$ diverges.
Let $f_n(x)=\sum_{j=1}^n(-1)^{j+1}/(x^2+\sqrt k\;)$ and $f(x)=\lim_{n\to \infty}.$ Now $1/(x^2+\sqrt k\;)$ is strictly decreasing in $ k$, so for any $x$ we have $|f_n(x)-f(x)|<1/(x^2+\sqrt {k+1}\;)\leq 1/\sqrt {k+1}.$ Since $1/\sqrt {k+1}\;$ is independent of $x,$ the convergence is uniform.