discrete mathematics - Proof by induction -inequality

Prove that $1 + \frac{1}{2}+ \frac{1}{3} + .... + 1/n < 2\sqrt{n}$ for $n \ge1$

Here's my attempt:

Base case:
$P(1): \frac{1}{1} \le 2\sqrt{1} = 1 \le 2$ (Base case true)

Assume $n = k$ for $k \ge1$ such that $1 + \frac{1}{2}+ \frac{1}{3} + .... + \frac{1}{k} < 2\sqrt{k}$

INDUCTION HYPOTHESIS: $1 + \frac{1}{2}+ \frac{1}{3} + .... + \frac{1}{k} < 2\sqrt{k}$

Now for P(k+1), we must show that $1 + \frac{1}{2}+ \frac{1}{3} + .... + \frac{1}{k+1}$ $< 2\sqrt{k+1}$

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$1 + \frac{1}{2}+ \frac{1}{3} + .... + \frac{1}{k} +\frac{1}{k+1}$ $< 2\sqrt{k+1}$

By our induction hypothesis: $2\sqrt{k} + \frac{1}{k+1} <2\sqrt{k+1}$

Now, I add $2\sqrt{k}$ to the right side.
$2\sqrt{k} + \frac{1}{k+1} <2\sqrt{k+1} + 2\sqrt{k}$ (the inequality is still true)

Then, substract $2\sqrt{k}$ from both sides,

Hence, $\frac{1}{k+1} <2\sqrt{k+1}$

QED

Also looking forward to see other ways to complete this induction! Thanks!

Answer

The problem within your proof is that you assume $1 + \frac{1}{2}+ \frac{1}{3} + .... + \frac{1}{k} +\frac{1}{k+1}< 2\sqrt{k+1}$ at the beginning but this is not allowed; you are trying to prove it but you assume it at first.

Another problem is that, when you apply the induction hypothesis, you are saying $A

To prove it you need to show $${1\over{k+1}}<2\sqrt{k+1}-2\sqrt{k}$$

which comes from

$${1\over{k+1}}={2\over{(k+1)+(k+1)}}<{2\over{\sqrt{k+1}+\sqrt{k}}}=2\sqrt{k+1}-2\sqrt{k}$$

Hence

$$1 + \frac{1}{2}+ \frac{1}{3} + .... + \frac{1}{k} +\frac{1}{k+1}<2\sqrt{k}+{1\over k+1}< 2\sqrt{k+1}$$