# Induction Problem

1. May 9, 2008

### dtl42

1. The problem statement, all variables and given/known data
Prove by induction. The sum from 1 to n of $$\frac{1}{\sqrt{i}}$$ $$\geq$$ $$\sqrt{n}$$.

2. Relevant equations
none.

3. 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

2. May 9, 2008

### 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?

3. May 10, 2008

### 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 $$\sum_{i=1}^n {\frac{1}{\sqrt{i}}} \geq \sqrt{n}$$

Last edited: May 10, 2008
4. May 10, 2008

### HallsofIvy

Staff Emeritus
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.