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

I understand that this is an induction question.

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

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?

• Assume for $k$ and prove it for $k+1$. – Berci May 6 '14 at 22:46
• You know the left hand side is no greater than $$2\sqrt{k} + \frac{1}{\sqrt{k+1}}.$$ – Daniel Fischer May 6 '14 at 22:47
• You must assume the case up to $k$, not $k+1$. The $k+1$ case is the one you need to prove. – Lucas Zanella May 6 '14 at 22:51
• @DanielFischer With that information, how can I prove the desired conclusion? I'm really lost, please elaborate. – user122661 May 7 '14 at 0:43
• If you can show that $$2\sqrt{k} + \frac{1}{\sqrt{k+1}} \leqslant 2\sqrt{k+1},$$ you're done. Showing that isn't too difficult. – Daniel Fischer May 7 '14 at 9:40

Hint We have that $(2 x^{1/2})'=x^{-1/2}$. Now, think about $$\int_1^n x^{-1/2}dx$$
Compare the area below the red curve ($y=1/\sqrt{x}$) and the blue curve from $x=0$ to $x=\sqrt{n}$.