The exact question:
Prove:
$\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}}\gt2(\sqrt{n+1}-1)$
I have looked at similar problems but still don't understand how to prove this inequality by induction. So far I have this:
Induction basis:
Let n=1
$\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}} = \frac{1}{\sqrt{1}} = 1 > 2(\sqrt{1+1}-1) = ~.828$
$1>.828$
So it proves the inequality true when n=1.
Now i really don't know how to continue even with all the examples i have browsed through. One of them i came across showed that the induction hypothesis should let P(n) equal the equation above and do something with P(n+1). I am not looking for the answer I just need help on how to continue with the problem. What other steps are necessary for me to complete this proof by induction.
Best Answer
Use this fact $$\sqrt{n+1} - \sqrt{n} = {1\over \sqrt{n} + \sqrt{n+1}}. $$ Now produce a bound and sum a telescoping sum.