Verify if $\sum(\sqrt{n+1}-\sqrt{n})$ is convergent or divergent 
Verify if the series $\sum a_n$ with $a_n=\sqrt{n+1}-\sqrt{n}$ is
convergent or divergent.

What I did is
$$a_n=(\sqrt{n+1}-\sqrt{n})\times\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt
{n+1}+\sqrt{n}}$$
$$=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
$$<\frac{1}{2\sqrt{n}}=b_n$$
Since $b_n$ is monotone decreasing and $b_n\rightarrow 0$ when $n\rightarrow \infty$ then $b_n$ is convergent.
Using the comparison test, we have that $0\leq a_n\leq b_n$. If $b_n$ is convergent then $a_n$ is convergent.
Is it right?
 A: In the same manner,
$$a_n>\frac1{\sqrt{n+1}+\sqrt{n+1}}=\frac12\frac1{\sqrt {n+1}}>\frac12\frac1{\sqrt{n+n}}=\frac1{2\sqrt2}\frac1{\sqrt n}=b_n$$
By the comparison test, we easily see that
$$\sum_{n=1}^\infty a_n>\sum_{n=1}^\infty b_n\to+\infty$$
which follows from the p-series.
A: The series diverges. This can be seen using the fact that it is a "telescoping series". 
The $k^\text{th}$ partial sum can be written as follows:
\begin{align}
S_k &= \displaystyle\sum_{n=1}^k \Big(\sqrt{n+1} - \sqrt{n}\Big)\\\\
&= \big(\sqrt{2} - \sqrt{1}\,\big) + \big(\sqrt{3} - \sqrt{2}\,\big) + \big(\sqrt{4} - \sqrt{3}\,\big) + \cdots +\big(\sqrt{k+1} - \sqrt{k}\,\big)\\\\
&=-\sqrt{1} + \big(\sqrt{2} -\sqrt{2}\,\big) + \big(\sqrt{3} -\sqrt{3}\,\big) + \big(\sqrt{4} - \cdots -\sqrt{k}\,\big) + \sqrt{k+1}\\\\
&=\sqrt{k+1} - 1\\
\end{align}
As $k$ goes to infinity, this diverges, so the infinite sum does not converge.
$$\displaystyle\sum_{n=1}^\infty \Big(\sqrt{n+1} - \sqrt{n}\Big) = \lim_{k\to\infty} S_k = \infty$$
A: $$\sqrt {n+1}-\sqrt {n}=$$
$$\sqrt {n}(\sqrt {1+\frac {1}{n}}-1)$$
$$\sim \sqrt {n}.\frac {1}{2n} $$
$$\sim \frac {1}{2\sqrt {n}} $$
the series diverges.
A: it is possible to show  that
$$\sum _{j=1}^{\infty } \left(\left(\sqrt{j+1}-\sqrt{j}\right)-\frac{\sqrt{\frac{1}{j}}}{2}\right)+2 \zeta \left(-\frac{1}{2}\right)+\frac{\zeta \left(\frac{1}{2}\right)}{2}=-\sqrt{2}$$ although mathematics numerically does not calculate it well, and therefore $$\sum _{j=1}^{\infty } \left(\sqrt{j+1}-\sqrt{j}\right)=-\sqrt{2}$$
