Correctness of the induction step Prove by induction that for all $ n \geq 2$ that the following:
$\sum_{i = 1}^n \frac{i}{i+1} < \frac{n^2}{n+1}$
Looking at the $n+1$ step, is it safe to assume the following ?
$\sum_{i = 1}^{n+1} \frac{i}{i+1} < \frac{(n+1)^2}{n+2} \iff \frac{n+1}{n+2} \leq \frac{(n+1)^2}{n+2} - \frac{n^2}{n+1}$
Assuming the inequality is true for $n$ we can derive that the inequality for $n+1$ stays true if the growth rate of the left side is lesser or equal to the growth rate of the right side.
We then solve the inequality on the right:
$(n+1)^2 \leq (n+1)^3 -n^2(n+2)$
$n^2+2n+1 \leq n^3+3n^2+3n+1 -n^3-2n^2$
$0 \leq n$
Since the questions assumes $ n \geq 2$  q.e.d.

Hint: You must show that $$\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{n}{n+1}+\frac{n+1}{n+2}<\frac{(n+1)^2}{n+2}$$ and this is (using your inequality)
$$\frac{n^2}{n+1}+\frac{n+1}{n+2}<\frac{(n+1)^2}{n+2}$$
And we get
$$\frac{(n+1)^2}{n+2}-\frac{n+1}{n+2}-\frac{n^2}{n+1}=\frac{n}{(n+1) (n+2)}$$
