Proving Induction: Sum 1 to n of 1/√i ≥ √n

[dtl42]

Homework Statement

Prove by induction. The sum from 1 to n of \frac{1}{\sqrt{i}} \geq \sqrt{n}.

Homework Equations

none.

The Attempt at a Solution

I verified the base case, and all of that, I just can't get to anything useful. I tried expanding the sum and adding the \frac{1}{\sqrt{n+1}} but it didn't come to anything useful.

Thanks

 

[Tedjn]

What have you gotten after adding the 1/\sqrt{n+1} ? What would you like the numerator of the fraction to be after combining?

 

[Kurret]

after the induction assumption you have to prove the inequality:
\sum_{i=1}^n {\frac{1}{\sqrt{i}}}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}
given that \sum_{i=1}^n {\frac{1}{\sqrt{i}}} \geq \sqrt{n}

 

[HallsofIvy]

Assuming \sum_{i=1}^n {\frac{1}{\sqrt{i}}} \geq \sqrt{n}. \sum_{i=1}^n {\frac{1}{\sqrt{i}}}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n}+ \frac{1}{\sqrt{n+1}}
Adding those, \sqrt{n}+ \frac{1}{\sqrt{n+1}}=\frac{\sqrt{n^2+ n}+1}{\sqrt{n+1}}
that's what you want if \sqrt{n^2+ n}+ 1\ge \sqrt{n+1} and that should be easy to prove.