# Absolute convergence tests

Hi I am interested in the following series:

$$\sum_{n=1}^{\infty}(-1)^{n}(\sqrt{n+3}-\sqrt{n})$$

I have been able to show that this series converges by proving the Leibnitz test. Does anyone know how to show that it converges absolutely? What tests are available for this?

Thanks for assistance.

• You could "rationalize the numerator" to get a form of the term in which it is clear whether it converges absolutely. Or you could use the Taylor expansion of the square root to obtain the magnitude of the terms. – Daniel Fischer Oct 5 '14 at 11:36
• @DanielFischer Yeah I have done that to show it converges. So if we take the absolute value we get $\frac{3}{\sqrt{n+3}+\sqrt{n}}$. How would the absolute convergence follow from this? Would you use some kind of comparison test? – user116403 Oct 5 '14 at 11:39
• It doesn't converge absolutely... – David Mitra Oct 5 '14 at 11:39
• You'd compare it to $\frac{c}{\sqrt{n}}$ for a suitable $c$. – Daniel Fischer Oct 5 '14 at 11:40
• And you are on the right path to prove that it does not converge, as it can be compared to $\frac{1}{\sqrt{n}}$ which diverge by the integral criteria. – PenasRaul Oct 5 '14 at 11:40

## 3 Answers

Notice that

$$\sqrt{n+3}-\sqrt n\ge \sqrt{n+1}-\sqrt n$$ and obviously the series $$\sum_{n\ge1}\sqrt{n+1}-\sqrt n$$ is divergent by telescoping. The given series isn't absolutely convergent.

• It seems that you have missed the factor $(-1)^n$... – Olivier Oloa Oct 5 '14 at 11:52
• Nope my answer is concerning the absolute convergence. – user63181 Oct 5 '14 at 11:54
• My mistake, you are completely right! – Olivier Oloa Oct 5 '14 at 11:57
• Convergence.

You may write \begin{align} \sum_{n=4}^{\infty}(-1)^{n}(\sqrt{n+3}-\sqrt{n}) &=\sum_{n=4}^{\infty}(-1)^{n}\sqrt{n}\left(\sqrt{1+\frac 3n}-1\right)\\\\ &=\sum_{n=4}^{\infty}(-1)^{n}\sqrt{n}\left(1+\frac{3}{2n}-\frac{9}{8n^2}+\mathcal{O}\left(\frac{1}{n^2}\right)-1\right)\\\\ &=\frac{3}{2}\sum_{n=4}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}-\frac{9}{8}\sum_{n=4}^{\infty}\frac{(-1)^{n}}{n\sqrt{n}}+\sum_{n=4}^{\infty}\mathcal{O}\left(\frac{1}{n\sqrt{n}}\right)\\\\ \end{align} and your initial series converges, being the sum of convergent series.

• Absolute convergence.

The same reasoning gives, for $N$ great, \begin{align} \sum_{n=4}^{N}(\sqrt{n+3}-\sqrt{n})&=\sum_{n=4}^{N}\sqrt{n}\left(\sqrt{1+\frac 3n}-1\right)\\\\ &=\sum_{n=4}^{N}\sqrt{n}\left(1+\frac{3}{2n}-\frac{9}{8n^2}+\mathcal{O}\left(\frac{1}{n^2}\right)-1\right)\\\\ &=\frac{3}{2}\sum_{n=4}^{N}\frac{1}{\sqrt{n}}-\frac{9}{8}\sum_{n=4}^{N}\frac{1}{n\sqrt{n}}+\sum_{n=4}^{N}\mathcal{O}\left(\frac{1}{n\sqrt{n}}\right) \end{align} and the series of absolute terms, being the sum of a divergent series and convergent series, is convergent and your initial series is absolutely divergent.

By the definition of absolutely convergent series you have to prove that $\sum_{n=1}^\infty|(-1)^n(\sqrt{n+3}-\sqrt{n})|$ doesn't converge $$S_k=\sum_{n=1}^k |(-1)^n(\sqrt{n+3}-\sqrt{n})|=\sum_{n=1}^k\sqrt{n+3}-\sqrt{n}=\sqrt{k+3}-1$$ $$\lim_{k\to\infty}S_k=\infty$$ From this we can say that the series isn't aboslutely convergent