Absolute convergence of the series $\sum_1^{\infty} {(-1)}^n(\sqrt{n+1} -\sqrt{n})$

$\sum_1^{\infty} {(-1)}^n(\sqrt{n+1} -\sqrt{n})$ I want to know if this series diverge or if it is absolutely convergent or conditionally convergent. I used Leibniz' Criterion of alternating series and I think that it converges conditionally, but I'm not sure if it is absolutely convergent. Can someone help me out please?

• $\sum_n(\sqrt{n+1}-\sqrt n)$ telescopes. – Lord Shark the Unknown Jun 2 '18 at 16:23

Hint: The absolute value of the general term has an equivalent: $$\sqrt{n+1}-\sqrt n=\frac 1{\sqrt{n+1}+\sqrt n}\sim_\infty \frac1{2\sqrt n}.$$ Does the latter converge?

$$\sqrt{n+1}-\sqrt{n}=\sqrt{n}\left(\sqrt{1+\frac{1}{n}}-1\right)$$ And $$\sqrt{1+\frac{1}{n}}\underset{(+\infty)}{=}1+\frac{1}{2n}+o\left(\frac{1}{n}\right)$$ So $$\sqrt{n+1}-\sqrt{n}\underset{(+\infty)}{=}\frac{1}{2\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)$$ The series $\displaystyle \sum \frac{1}{\sqrt{n}}$ diverges, so it does not converge absolutely.

$\begin{array}\\ \text{If}\\ s(m) &=\sum_{n=1}^{m} {(-1)}^n(\sqrt{n+1} -\sqrt{n})\\ \text{then}\\ s(2m) &=\sum_{n=1}^{2m} {(-1)}^n(\sqrt{n+1} -\sqrt{n})\\ &=\sum_{n=1}^m (-(\sqrt{2n}-\sqrt{2n-1})+(\sqrt{2n+1}-\sqrt{2n}))\\ &=\sum_{n=1}^m (\dfrac{-1}{\sqrt{2n}+\sqrt{2n-1}}+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n}})\\ &=\sum_{n=1}^m \dfrac{-(\sqrt{2n+1}+\sqrt{2n})+\sqrt{2n}+\sqrt{2n-1}}{(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})}\\ &=\sum_{n=1}^m \dfrac{-\sqrt{2n+1}+\sqrt{2n-1}}{(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})}\\ &=\sum_{n=1}^m \dfrac{-2}{(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})}\\ \text{so}\\ s(2m) &>-2\sum_{n=1}^m \dfrac{1}{8(2n-1)^{3/2}}\\ &=-\frac14\sum_{n=1}^m \dfrac{1}{(2n-1)^{3/2}}\\ \text{and}\\ s(2m) &<-2\sum_{n=1}^m \dfrac{1}{8(2n+1)^{3/2}}\\ &=-\frac14\sum_{n=1}^m \dfrac{1}{(2n+1)^{3/2}}\\ \end{array}$

and both bounds converge.

Therefore the whole sum converges since $(\sqrt{n+1}-\sqrt{n}) =\dfrac1{\sqrt{n+1}+\sqrt{n}} \to 0$.

Note: Wolfy says that the sum of the first 1,000,000 terms is about -0.23954, validates my algebra, and says that

$(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})\\ =8(2n)^{3/2} -\dfrac{5}{2\sqrt{2n}} -\dfrac{17}{32(2n)^{5/2}} +O(n^{-9/2})$.