Determine the convergence of $\sum (\sqrt{n+1}) -(\sqrt{n})$

Question

I tried using the conjugate of $\sum (\sqrt{n+1}) -(\sqrt{n})$, which leads to

$\sum \frac{1}{(\sqrt{n+1})+(\sqrt{n})}$,

and tried to comparing it with

$\sum \frac{1}{(\sqrt{n+1})+(\sqrt{n})}\geq \sum \frac{1}{2(\sqrt{n+1})}$,

which I do not know the further steps to this. Can anyone help me out?

Answer 1

Since $(\forall n\in\mathbb{N}):\sqrt{n+1}\leqslant n+1$, you have that $(\forall n\in\mathbb{N}):\frac1{2\sqrt{n+1}}\geqslant\frac1{2(n+1)}$. So…

Answer 2

You are almost done, simply note that

$$\frac{1}{2\sqrt{n+1}}\ge \frac1{2(n+1)}$$

which diverges by harmonic series.

Answer 3

Even easier(without resorting to the harmonic series): note that $1 \geq \frac{1}{\sqrt{n}}, \frac{1}{2} \geq \frac{1}{\sqrt{n}}, ..., \frac{1}{\sqrt{n-1}} \geq \frac{1}{\sqrt{n}} and \frac{1}{\sqrt{n}} \geq \frac{1}{\sqrt{n}}$, which when added up give $\sum_{i=1}^{n}\frac{1}{\sqrt{i}} \geq n\frac{1}{\sqrt{n}}=\sqrt{n}$ which clearly tends to infinity

Answer 4

$$\sum_{i=1}^n\sqrt{i+1}-\sqrt i=\sqrt{n+1}-1\to \infty$$therefore the series is divergent.