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

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