Mathematical Induction: how do I prove inequality

Question

I got stuck solving this mathematical induction. I know how to solve basic inquality, but not sure what to do with sum in the inequality.

√n < √(1/1) + √(1/2) + ··· +√(1/n), for all integers n ≥ 2

What I did:

1. P(2) => √(1/1) + √(1/2) > √(2). True

2. P(k) => √(1/1) + √(1/2) + ··· + √(1/k) > √k

3. P(k+1) => √(1/1) + √(1/2) + ··· + √( 1/(k+1) ) > √(k+1)

√(1/1) + √(1/2) + ··· +√(1/k) + √( 1/(k+1) ) > √(k+1)

And we know that √(1/1) + √(1/2) + ··· +√(1/k) > √k

Therefore,

√k + √( 1/(k+1) ) > √(k+1)

But is it right to do so? Don't I violate anything?

Answer 1

You are doing right. Just to finish, you want to show that:

$$\sqrt{k}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$$

but

$$\sqrt{k}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1} \Leftrightarrow \sqrt{k(k+1)}+1>k+1 \Leftrightarrow$$

$$\Leftrightarrow \sqrt{k(k+1)}>k \Leftrightarrow \sqrt{k+1}>\sqrt{k}$$

and the last inequality is obvious.