# Show that $\sum_{j=1}^{\infty}\frac{\sqrt{j+1}-\sqrt{j}}{j+1}$ is convergent or divergent.

How would I show that the following series converges or diverges.

$\sum_{j=1}^{\infty}\frac{\sqrt{j+1}-\sqrt{j}}{j+1}$

I am not sure what is the best test to show that it converges. Apparently it converges.

• Start by rationalizing the numerator using $(x-y)(x+y)=x^2-y^2$ to get $\sqrt{j+1}-\sqrt{j} = \frac{1}{\sqrt{j+1}+\sqrt{j}}$. – Winther Oct 11 '15 at 17:54
• @Winther I think that would almost do as an answer. – mickep Oct 11 '15 at 17:58

You may write $$\sum_{j=1}^{\infty}\frac{\sqrt{j+1}-\sqrt{j}}{j+1}=\sum_{j=1}^{\infty}\frac1{(j+1){(\sqrt{j+1}+\sqrt{j}})}$$ and observe that, as $j \to \infty$, $$\frac1{(j+1){(\sqrt{j+1}+\sqrt{j}})} \sim \frac1{2j^{3/2}}$$ leading to the convergence of the initial series.
$$\frac{1}{\sqrt{j}}-\frac{1}{\sqrt{j+1}} = \frac{\sqrt{j+1}-\sqrt{j}}{\sqrt{j^2+j}}\geq\frac{\sqrt{j+1}-\sqrt{j}}{j+1}\geq 0 \tag{1}$$ hence: $$0 \leq \sum_{j\geq 1}\frac{\sqrt{j+1}-\sqrt{j}}{j+1}\leq\sum_{j\geq 1}\left(\frac{1}{\sqrt{j}}-\frac{1}{\sqrt{j+1}}\right)=\color{red}{1}.\tag{2}$$