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.