Is this string  ascending or descending? I need help in checking that string is ascending or descending and I have problems with it.
so it's my "$a_n$"
$$
\sqrt {n+5} - \sqrt{n}
$$
and I need to use $a_{n+1} - a_n$ or $\frac {a_n+1}{a_n}$.
EDIT:
it would be something like that 
$$
\sqrt {n+6} - \sqrt{n+1} -\sqrt{n+5}+\sqrt{n}
$$
and what's next?
 A: \begin{align*}
a_n &= \sqrt{n+5}-\sqrt{n} = \left(\sqrt{n+5}-\sqrt{n}\right)\frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+5}+\sqrt{n}} \\
&= \frac{\left(\sqrt{n+5}-\sqrt{n}\right)\left(\sqrt{n+5}+\sqrt{n}\right)}{\sqrt{n+5}+\sqrt{n}} \\
&= \frac{n+5 - n}{\sqrt{n+5}+\sqrt{n}} \\
&= \frac{5}{\sqrt{n+5}+\sqrt{n}} \\
\end{align*}
This is clearly a decreasing sequence. As $n$ increases, the denominator increases so the sequence as a whole decreases.

If you insist on using $\frac{a_{n+1}}{a_n}$, use the same approach above to show that:
\begin{align*}
\frac{a_{n+1}}{a_n} &= \frac{\sqrt{n+6}-\sqrt{n+1}}{\sqrt{n+5}-\sqrt{n}} \\
&= \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}}
\end{align*}
Now, notice that:
$$
n+1 \gt n \Rightarrow \sqrt{n+1} \gt \sqrt{n}
$$
And:
$$
n+6 \gt n+5 \Rightarrow \sqrt{n+6} \gt \sqrt{n+5}
$$
Add the inequalities side by side to get:
$$
\sqrt{n+6} + \sqrt{n+1} \gt \sqrt{n+5} + \sqrt{n}
$$
Divide both sides by $\sqrt{n+6} + \sqrt{n+1}$ to get:
$$
\frac{a_{n+1}}{a_n} = \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}} < 1
$$
A: How about calculus?
Differentiate the function $f(x)=\sqrt{x+5}-\sqrt{x}$. We get $f'(x) = \frac12(\sqrt{\frac{1}{x+5}}-\sqrt{\frac{1}{x}})$. This is negative for all $x\ge 0$ because $\sqrt{x}$ is strictly increasing. Therefore our $f$ is strictly decreasing, so the sequence formed by taking its values at discrete $x$s is decreasing too.
A: $(\sqrt {n+5} - \sqrt{n})\times(\sqrt {n+5} + \sqrt{n})={n+5} - {n} =5$ which is constant.  
But $\sqrt {n+5} + \sqrt{n}$ is increasing with positive $n$ so $\sqrt {n+5} - \sqrt{n}$ is decreasing.
