Proving that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$

Question

How do I prove that:

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$$

Do I use induction?

Answer 1

Prove the following claim using induction on $n$: $$\sum_{k=1}^n \dfrac1{\sqrt{k}} < 2 \sqrt{n}$$

In the induction, you will essentially need to show that $$2\sqrt{n} +\dfrac1{\sqrt{n+1}} < 2 \sqrt{n+1} \tag{$\star$}$$

To prove $(\star)$, note that $$\sqrt{n} < \sqrt{n+1} \implies \sqrt{n} + \sqrt{n+1} <2 \sqrt{n+1} \implies \dfrac1{\sqrt{n+1}} < \dfrac2{\sqrt{n} + \sqrt{n+1}}$$ Multiplying and divding the right hand side by $(\sqrt{n+1} - \sqrt{n})$, we get $$\dfrac1{\sqrt{n+1}} < \dfrac2{\sqrt{n} + \sqrt{n+1}}\cdot \dfrac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = 2({\sqrt{n+1} - \sqrt{n}})$$ which gives us $(\star)$.

Answer 2

You can use integral:

$$\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +\dots +\frac { 1 }{ \sqrt { 100 } } <\int _{ 0 }^{ 100 }{ \frac { 1 }{ \sqrt { x } } } dx=20$$

You can imagine approximating the integral with rectangles of side $\frac { 1 }{ \sqrt { n } }$ and $1$, will give less area than the integral because of the behaviour of the curve.