Show that $\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \sqrt {n + 1}  - \sqrt n$ where $n \ge 0,{\text{ }}n \in \mathbb{Z}$.

Hence show that $\sqrt 2  - 1 < \frac{1}{{\sqrt 2 }}$.

Prove, by mathematical induction, that $\sum\limits_{r = 1}^{r = n} {\frac{1}{{\sqrt r }} > \sqrt n }$ for $n \ge 2,{\text{ }}n \in \mathbb{Z}$.

$\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \frac{1}{{\sqrt n  + \sqrt {n + 1} }} \times \frac{{\sqrt {n + 1}  - \sqrt n }}{{\sqrt {n + 1}  - \sqrt n }}$     M1

$= \frac{{\sqrt {n + 1}  - \sqrt n }}{{(n + 1) - n}}$     A1

$= \sqrt {n + 1}  - \sqrt n$     AG

[2 marks]

$\sqrt 2  - 1 = \frac{1}{{\sqrt 2  + \sqrt 1 }}$     A2

$< \frac{1}{{\sqrt 2 }}$     AG

[2 marks]

consider the case $n = 2$: required to prove that $1 + \frac{1}{{\sqrt 2 }} > \sqrt 2$     M1

from part (b) $\frac{1}{{\sqrt 2 }} > \sqrt 2  - 1$

hence $1 + \frac{1}{{\sqrt 2 }} > \sqrt 2$ is true for $n = 2$     A1

now assume true for $n = k:\sum\limits_{r = 1}^{r = k} {\frac{1}{{\sqrt r }} > \sqrt k }$     M1

$\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} > \sqrt k$

attempt to prove true for $n = k + 1:\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}$     (M1)

from assumption, we have that $\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt k  + \frac{1}{{\sqrt {k + 1} }}$     M1

so attempt to show that $\sqrt k  + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}$     (M1)

EITHER

$\frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}  - \sqrt k$     A1

$\frac{1}{{\sqrt {k + 1} }} > \frac{1}{{\sqrt k  + \sqrt {k + 1} }}$, (from part a), which is true     A1

OR

$\sqrt k  + \frac{1}{{\sqrt {k + 1} }} = \frac{{\sqrt {k + 1} \sqrt k  + 1}}{{\sqrt {k + 1} }}$     A1

$> \frac{{\sqrt k \sqrt k  + 1}}{{\sqrt {k + 1} }} = \sqrt {k + 1}$     A1

THEN

so true for $n = 2$ and $n = k$ true $\Rightarrow n = k + 1$ true. Hence true for all $n \ge 2$     R1

Note:     Award R1 only if all previous M marks have been awarded.

[9 marks]

Total [13 marks]