<s>[INST] 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.
 [/INST] No, it is not safe to assume that. Since this is supposed to be an induction proof, what you should assume is that $\sum_{i=1}^n\frac i{i+1}<\frac{n^2}{n+1}$. Then$$\sum_{i=1}^{n+1}\frac i{i+1}=\left(\sum_{i=1}^n\frac i{i+1}\right)+\frac{n+1}{n+2}<\frac{n^2}{n+1}+\frac{n+1}{n+2}.$$So, all that remains to be proved is that$$\frac{n^2}{n+1}+\frac{n+1}{n+2}\leqslant\frac{(n+1)^2}{n+2}.$$
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