Prove that for every integer $n \ge 1$, $1 + \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+ ... +\frac{1}{\sqrt{n}}\le 2\sqrt{n}$

Question

I understand that this is an induction question.

I start with the base case (n=1):

$$1 < 2 \tag{That works!}$$

Induction step: Assume the statement works for all $n = k$, Prove for all $n = k+1$

Assume $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}+ ... +\frac{1}{\sqrt{k+1}}\le 2\sqrt{k+1}$

I'm a bit confused as to where to go next, may I please have some assistance?

Answer 1

Hint We have that $(2 x^{1/2})'=x^{-1/2}$. Now, think about $$\int_1^n x^{-1/2}dx$$

Answer 2

Compare the area below the red curve ($y=1/\sqrt{x}$) and the blue curve from $x=0$ to $x=\sqrt{n}$.