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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?

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  • $\begingroup$ Assume for $k$ and prove it for $k+1$. $\endgroup$ – Berci May 6 '14 at 22:46
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    $\begingroup$ You know the left hand side is no greater than $$2\sqrt{k} + \frac{1}{\sqrt{k+1}}.$$ $\endgroup$ – Daniel Fischer May 6 '14 at 22:47
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    $\begingroup$ You must assume the case up to $k$, not $k+1$. The $k+1$ case is the one you need to prove. $\endgroup$ – Lucas Zanella May 6 '14 at 22:51
  • $\begingroup$ @DanielFischer With that information, how can I prove the desired conclusion? I'm really lost, please elaborate. $\endgroup$ – user122661 May 7 '14 at 0:43
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer May 7 '14 at 9:40
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Hint We have that $(2 x^{1/2})'=x^{-1/2}$. Now, think about $$\int_1^n x^{-1/2}dx$$

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  • $\begingroup$ Thanks for the response, however we haven't ever used integration in this course (in fact, I haven't learned it at all). $\endgroup$ – user122661 May 7 '14 at 0:45
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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

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  • $\begingroup$ Did you see my answer? =) $\endgroup$ – Pedro Tamaroff May 6 '14 at 23:49

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