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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?

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What do you mean? The former term is clearly greater than the latter. –  M.B. Jun 10 '12 at 21:26
    
Yes but it would be something like that $$ \sqrt {n+6} - \sqrt{n+1} -\sqrt{n+5}+\sqrt{n} $$ and what's next? –  Sebastian Stasiak Jun 10 '12 at 21:29

3 Answers 3

\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:

$$ \frac{a_{n+1}}{a_n} = \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}} < 1 $$

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The OP needs to use $a_{n+1} - a_n$ or $\frac {a_n+1}{a_n}$. –  Gigili Jun 10 '12 at 21:41
    
@Gigili - Added a proof using $\frac{a_{n+1}}{a_n}$. –  Ayman Hourieh Jun 10 '12 at 21:55
    
Great, Plus-one'd. –  Gigili Jun 10 '12 at 21:57

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.

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$(\sqrt {n+5} - \sqrt{n})\times(\sqrt {n+5} + \sqrt{n})={n+5} - {n} =5$ which is constant.

But $\sqrt {n+5} + \sqrt{n}$ is increasing with positive $n$ so $\sqrt {n+5} - \sqrt{n}$ is decreasing.

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Much the same as Ayman Hourieh's answer, but perhaps easier to understand for some readers –  Henry Jun 10 '12 at 21:32

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