$f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$, $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$

Question

Given $f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$
Evaluate $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$

I have simplified the given to $f(n)=\sqrt{n}+\sqrt{n+1}$ but I am still not sure of how to solve this.

Answer 1

The hint.

Use $$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$$ and the telescopic sum.

Finally, we obtain $$\sqrt{100}-1=9.$$

Answer 2

Very simple Hint: Rationalize the denominator

$$\frac1{f(x)}=\frac1{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})}=\sqrt{n+1}-\sqrt{n}$$