This is a solution that my friend has suggested. Basically it does a lot of induction to come up with the proof! (Isn't that what mathematical induction is all about?)
I will continue from equation (3) in my original message, which is: \sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} >...
That's a lot to process. Estimating asymptotes is something I am not so confident about. I will give it a try. But I do agree that the professor wants us to use inductive reasoning to solve the problem, so resorting to graphical analysis may not satisfy him. However, this professor who taught us...
I think I have come up with a reasonable solution:
Continuing from my last equation, which is: (k+1)\sqrt{k} + \sqrt{k+2} > (k+1)\sqrt{k+1}
I rearrange it to form: 0 > (k+1)\sqrt{k+1} - (k+1)\sqrt{k} - \sqrt{k+2}
which is: (k+1)\sqrt{k+1} - (k+1)\sqrt{k} - \sqrt{k+2} < 0
or...
How do I get rid of them?
If you mean to do squaring both sides, well there is a + sign on LHS, so it's going to get longer and still have a square root.
Also if I move the non-square root terms to the right side and square the equation again, it's going to get much longer and more...
I guess so. For \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}, how do I prove that the LHS is greater than the RHS for all positive integers?
(PS: I can use calculus to solve the problem. Could you guide me how to use calculus to solve inequality?)
Homework Statement
\forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}
Homework Equations
I have to prove the above via mathematical induction.
The Attempt at a Solution
I did the base case, n = 1 and found it true for the base case.
Then I assumed that...