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?
On
The idea of the author is that if you "stop" the sum at the $n$'th term you get this artificial bound of $2-$something. Then you can show by induction that this holds for every $n$. Having proven this, the assertion follows by taking the limit as $n\rightarrow \infty$ since the bound will always be less than $2$.
Long story short, the author creates a bound which is provable by induction, that's where the $\frac{2}{\sqrt{n+1}}$ comes 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?