Solution check to Absolute convergence test

Question

By the absolute convergence test, the series $$\sum \frac{(-1)^{n}}{\sqrt{n+1}+\sqrt{n}} =\sum a_{n}$$ diverges since $$\sum \frac{(-1)^{n}}{\sqrt{n+1}+\sqrt{n}} < \sum \frac{1}{n^{\frac{1}{2}}}$$ and by the p-series $$\sum \frac{1}{n^{\frac{1}{2}}}$$
diverges for $$p<1$$ and by the direct comparison test, since $$b_{n}= \sum \frac{1}{n^{\frac{1}{2}}}$$ diverges, so does $$\sum \frac{(-1)^{n}}{\sqrt{n+1}+\sqrt{n}}$$ But by the alternating series test, $$\sum \frac{(-1)^{n}}{\sqrt{n+1}+\sqrt{n}} =\sum a_{n}$$ converges and so the series $$\sum \frac{(-1)^{n}}{\sqrt{n+1}+\sqrt{n}} =\sum a_{n}$$ is conditionally convergent.

Does this look right?

Answer 1

I assume you mean to show that the series is not absolutely convergent in the first part. One minor problem is that you need to write $\sum\frac1{\sqrt{n+1}+\sqrt n}$ since you're taking the absolute value and because the comparison test only works for series of non-negative terms.

More serious that is that you need to show that the terms of a series are greater than the terms of a divergent series in order to conclude that the original series is divergent. Your inequality is going the wrong way.