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. 
 A: Use this fact
$$\sqrt{n+1} - \sqrt{n} = {1\over \sqrt{n} + \sqrt{n+1}}. $$
Now produce a bound and sum a telescoping sum.
A: First you have to establish your statement of $P(n)$. Here the statement should be:
$$ P(n) \,\, : \,\, \displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}}\gt2(\sqrt{n+1}-1)$$
Now you go into the induction part. Equipped with the hypothesis that $P(n)$ is true, you need to prove that $P(n+1)$ is also true.
$$ P(n+1) \,\, : \,\, \displaystyle\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}}\gt2(\sqrt{(n+1)+1}-1)$$
A: The induction step assumes
$$\sum_{k = 1}^n \frac{1}{\sqrt{k}} > 2\sqrt{n + 1} - 2.$$
Using it, it follows that
\begin{align}
& \sum_{k = 1}^{n + 1} \frac{1}{\sqrt{k}} \\
= & \sum_{k = 1}^n \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{n + 1}} \\
> &2\sqrt{n + 1} - 2 + \frac{1}{\sqrt{n + 1}} \\
= & 2 \sqrt{n + 2} - 2 + \left[2(\sqrt{n + 1} - \sqrt{n + 2}) + \frac{1}{\sqrt{n + 1}}\right] \\
= & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}\right] \quad \text{multiply conjugate}\\
> & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{2\sqrt{n + 1}}\right] \\
= & 2 \sqrt{(n + 1) + 1} - 2,
\end{align}
finishes the induction step.
A: 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
