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?

You know the left hand side is no greater than $$2\sqrt{k} + \frac{1}{\sqrt{k+1}}.$$ — Daniel Fischer, May 06 '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. — PPP, May 06 '14 at 22:51
@DanielFischer With that information, how can I prove the desired conclusion? I'm really lost, please elaborate. — user122661, May 07 '14 at 00:43
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 07 '14 at 09:40
http://math.stackexchange.com/questions/768916/prove-that-2-sqrt-n-ge-1-sqrt11-sqrt2-cdots1-sqrt-n-ge2-sqrtn1-1/769181#769181 — medicu, May 07 '14 at 11:38

score 4 · Answer 1 · answered May 06 '14 at 22:46

4

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

answered May 06 '14 at 22:46

Pedro

Thanks for the response, however we haven't ever used integration in this course (in fact, I haven't learned it at all). – user122661 May 07 '14 at 00:45

score 0 · Answer 2 · answered May 06 '14 at 23:46

0

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

enter image description here

answered May 06 '14 at 23:46

2 Answers2