Show that $\frac{1}{\sqrt{r+1}+\sqrt{r}}=\sqrt{r+1}-\sqrt{r}$ for all $r\in {{\mathbb{Z}}^{+}}$. Hen

Question

Show that $\frac{1}{\sqrt{r+1}+\sqrt{r}}=\sqrt{r+1}-\sqrt{r}$ for all $r\in {{\mathbb{Z}}^{+}}$. Hence find $\sum\limits_{r=1}^{n}{\frac{1}{\sqrt{r+1}+\sqrt{r}}}$. Deduce that $\sum\limits_{r=1}^{n}{\frac{1}{\sqrt{r}}>2\left( \sqrt{n+1}-1 \right)}$.

Accepted Answer

(i) shown; (ii) $\sqrt{n+1}-1$; (iii) shown

2013 JJC P1 Q3

2013 JJC P1 Q3

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