# Induction Problem

## Homework Statement

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

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

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

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:
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.