agent007bond
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- Feb 22, 2010
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I have to prove the following via mathematical induction, \(\displaystyle \forall n \in Z^+\)
\(\displaystyle \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}\)
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: \(\displaystyle \sum_{i=1}^k \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2}\) ... (1)
Then, when n = k+1, \(\displaystyle \sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k+1}}{2}\) ... (2)
The LHS is equivalent to: \(\displaystyle \sum_{i=1}^k \frac{\sqrt{i+1}}{2i} + \frac{\sqrt{k+2}}{2k+2}\)
Comparing with equation (1) I can write the following: \(\displaystyle \sum_{i=1}^k \frac{\sqrt{i+1}}{2i} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2}\)
Hence, \(\displaystyle \sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2}\) ... (3)
According to our lecturer, from (2) and (3) we have: \(\displaystyle \sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}\)
Therefore to prove (2), we need to prove the inequality: \(\displaystyle \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}\)
Simplifying it, I can come up with: \(\displaystyle (k+1)\sqrt{k} + \sqrt{k+2} > (k+1)\sqrt{k+1}\)
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?!
\(\displaystyle \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}\)
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: \(\displaystyle \sum_{i=1}^k \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2}\) ... (1)
Then, when n = k+1, \(\displaystyle \sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k+1}}{2}\) ... (2)
The LHS is equivalent to: \(\displaystyle \sum_{i=1}^k \frac{\sqrt{i+1}}{2i} + \frac{\sqrt{k+2}}{2k+2}\)
Comparing with equation (1) I can write the following: \(\displaystyle \sum_{i=1}^k \frac{\sqrt{i+1}}{2i} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2}\)
Hence, \(\displaystyle \sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2}\) ... (3)
According to our lecturer, from (2) and (3) we have: \(\displaystyle \sum_{i=1}^{k+1} \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}\)
Therefore to prove (2), we need to prove the inequality: \(\displaystyle \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}\)
Simplifying it, I can come up with: \(\displaystyle (k+1)\sqrt{k} + \sqrt{k+2} > (k+1)\sqrt{k+1}\)
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?!