- #1
agent007bond
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Homework Statement
[tex]\forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}[/tex]
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 the proposition is true for n = k: [tex]\sum_{i=1}^k \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2}[/tex] ... (1)
Then, when n = k+1, [tex]\sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k+1}}{2}[/tex] ... (2)
The LHS is equivalent to: [tex]\sum_{i=1}^k \frac{\sqrt{i+1}}{2i} + \frac{\sqrt{k+2}}{2k+2}[/tex]
Comparing with equation (1) I can write the following: [tex]\sum_{i=1}^k \frac{\sqrt{i+1}}{2i} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2}[/tex]
Hence, [tex]\sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2}[/tex] ... (3)
According to our lecturer, from (2) and (3) we have: [tex]\sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}[/tex]
Therefore to prove (2), we need to prove the inequality: [tex]\frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}[/tex]
Simplifying it, I can come up with: [tex](k+1)\sqrt{k} + \sqrt{k+2} > (k+1)\sqrt{k+1}[/tex]
How do I prove this inequality (in order to prove the proposition)? Many of my friends in school are also as stuck as I am! Could someone help me?!