Use induction to prove that $ 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} .... + \frac {1}{\sqrt{n}} < 2\sqrt{n}$

Use induction to prove that $ 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} ... + \frac {1}{\sqrt{n}} < 2\sqrt{n} $

My attempt was as follows:

Lets assume the inequality is true for n = k

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

$ => S_k < 2\sqrt{k} $

$ => S_k < 2\sqrt{k + 1} $

We need to prove that

$ => S_k + \frac {1}{\sqrt{k+1}} < 2\sqrt{k + 1} $

$ => \frac {1}{\sqrt{k+1}} < 2\sqrt{k + 1} - S_k $

Now I don't know where to go from here please help

Solution 1:

$$2\sqrt{k} + \frac {1}{\sqrt{k+1}} = \frac {2\sqrt{k^2+k}+1}{\sqrt{k+1}} < \frac {2\sqrt{k^2+k+\dfrac14}+1}{\sqrt{k+1}} = \frac{2\left(k+\dfrac12\right)+1}{\sqrt{k + 1}}= 2\sqrt{k + 1}.$$