Proving that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$ How am I suppose to prove that:
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$$
Do I use the way like how we count $1+2+ \cdots+100$ to estimate?
So $1/5050 \lt 20$, implying that it is indeed less than $20$?

Start from $\sqrt{n}-\sqrt{n-1}$. Multiply top and "bottom" by $\sqrt{n}+\sqrt{n-1}$. 
We get $\dfrac{1}{\sqrt{n}+\sqrt{n-1}}$, which is $\gt \dfrac{1}{2\sqrt{n}}$. It follows that $\dfrac{1}{\sqrt{n}}\lt 2(\sqrt{n}-\sqrt{n-1})$.
Add up, $n=1$ to $n=100$. We get 
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{100}}\lt 2(\sqrt{1}-\sqrt{0})+2(\sqrt{2}-\sqrt{1})+\cdots +2(\sqrt{100}-\sqrt{99}).$$
Observe the mass cancellation on the right: it collapses to $20$.  
