discrete mathematics - Proving a summation inequality with induction

The exact question:

Prove:

$\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}}\gt2(\sqrt{n+1}-1)$

I have looked at similar problems but still don't understand how to prove this inequality by induction. So far I have this:

Induction basis:

Let n=1

$\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}} = \frac{1}{\sqrt{1}} = 1 > 2(\sqrt{1+1}-1) = ~.828$

$1>.828$

So it proves the inequality true when n=1.

Now i really don't know how to continue even with all the examples i have browsed through. One of them i came across showed that the induction hypothesis should let P(n) equal the equation above and do something with P(n+1). I am not looking for the answer I just need help on how to continue with the problem. What other steps are necessary for me to complete this proof by induction.

Answer

Inductive steps:

Assume the inequality is true for $n=N$.

so $\displaystyle\sum_{k=1}^N \frac{1}{\sqrt{k}}\gt2(\sqrt{N+1}-1)$

Now if $n=N+1$,

$\displaystyle\sum_{k=1}^{N+1} \frac{1}{\sqrt{k}} =\displaystyle\sum_{k=1}^N \frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{N+1}} >2(\sqrt{N+1}-1) + \frac{1}{\sqrt{N+1}} $

... (I left the part here for you to figure out)

$>2(\sqrt{N+2}-1)$

Hint: you want to prove that $\frac{1}{\sqrt{n}} \geq 2(\sqrt{N+2}-\sqrt{N+1})$ by multiplying by the conjugate term