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?

Hint: Note that $$\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}<\frac{1}{\sqrt{k+1}}$$
