$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{x^2+\sqrt{k}}$

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?

  • $\begingroup$ If I were grading this argument, I'd want to see an upper bound independent of $x$ in order to feel that "uniform convergence" had been justified. :) For absolute convergence, fix $x$ and think about how rapidly (or slowly) $\frac{1}{x^{2} + \sqrt{k}}$ decays as $k \to \infty$. $\endgroup$ – Andrew D. Hwang May 21 '17 at 11:10
  • $\begingroup$ You want to say $|R_n(x)| \leq \frac{1}{x^2+\sqrt{n+1}}\le \frac{1}{\sqrt{n+1}} \to 0.$ $\endgroup$ – zhw. May 21 '17 at 15:29
  • $\begingroup$ But if I keep $x^2$ term, where is the problem? The inequality still valid for any x. $\endgroup$ – Nour May 21 '17 at 16:14
  • $\begingroup$ You want a bound independent of $x$ $\endgroup$ – zhw. May 21 '17 at 16:48

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.

  • $\begingroup$ I added a note about uniform convergence. $\endgroup$ – DanielWainfleet May 21 '17 at 17:45

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