Prove $ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2$ 
For any positive integer $n$ prove by induction that:
  $$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2.$$  

The author  says that   it  is  sufficient to prove  that
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2-\frac{2}{\sqrt{n+1}}.$$
Why? Where this $\frac{2}{\sqrt{n+1}}$ term come from?

The stronger inequality is easier to be proved by using induction than the original one. This is another example: in order to prove
$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} < 2$$  show that
$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} \le 2-\dfrac{1}{n}.$$
See Induction on inequalities: $\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}<2$?
In our case, at the inductive step, it suffices to show that
$$2-\frac{2}{\sqrt{n+1}}+\frac{1}{(n+2)\sqrt{n+1}}<2-\frac{2}{\sqrt{n+2}}.$$
Can you take it from here?
