Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: Let the given statement be $p(n)$ .
1.\begin{align*} \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} & =\frac{\sqrt 2 +1}{\sqrt 2} \\ 
2 &< \sqrt 2 +1 \\
\sqrt 2 &< \frac{\sqrt 2 +1}{\sqrt 2}=\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}  \end{align*}
Hence, $p(2)$ is true.
2.For an arbitrary integer $k \ge 2$, suppose $p(k)$ is true.
That is, $$\sqrt k < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}$$
Then we must show that $p(k+1)$ is true.
We're going to show that $$\sqrt {k+1} < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt {k+1}}$$
I'm stuck on this step. I can't develop it further. How can I complete this proof?
 A: $$\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt k+1} > \sqrt{k} + \frac{1}{\sqrt {k+1}} =
\frac{\sqrt{k(k+1)} + 1}{\sqrt{k+1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}} = \sqrt{k+1}$$
A: Hint: Note that $$\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}<\frac{1}{\sqrt{k+1}}$$
A: For the last induction step
$$\sqrt{k^2 + k}+1> k +1 \implies \frac{\sqrt{k^2 + k}}{\sqrt{k+1}}+ \frac1{\sqrt{k+1}}> \sqrt{k+1}\implies  \sqrt{k} +\frac1{\sqrt{k+1}}> \sqrt{k+1}.$$
Hence
$$\sum_{j=1}^{k} \frac1{\sqrt{j}} > \sqrt{k} \implies \sum_{j=1}^{k+1} \frac1{\sqrt{j}} > \sqrt{k} + \frac1{\sqrt{k+1}} > \sqrt{k+1}.$$
For a simpler proof
$$\sum_{k=1}^n \frac1{\sqrt{k}}> \frac{n}{\sqrt{n}}= \sqrt{n}.$$
A: For the inductive step, it suffices to show $\sqrt{n+1} - \sqrt{n} < \frac {1}{\sqrt{n+1}} = \frac{\sqrt{n+1}}{n+1}$ since that implies the inequality gets stronger as $n$ increases. This is clear since 
$$\sqrt{n+1} - \sqrt{n} =   \frac{\sqrt{n+1}}{n+1} + \frac{(-\sqrt{n}\sqrt{n+1})(\sqrt{n+1} - \sqrt{n})}{n+1}$$
The right addend on the right equation is clearly negative. 
