Correctness of the induction step

Question

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.

Answer 1

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$$\frac{n^2}{n+1}+\frac{n+1}{n+2}\leqslant\frac{(n+1)^2}{n+2}.$$

Answer 2

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 $$\frac{(n+1)^2}{n+2}-\frac{n+1}{n+2}-\frac{n^2}{n+1}=\frac{n}{(n+1) (n+2)}$$