Derivation of inequality $\frac{1}{(n+2)^2}<\frac{1}{n+1}-\frac{1}{n+2}$ in proof?

Question

I don't understand how it went from line four to five in the proof? Do you need to use induction? We haven't covered it yet.

Answer 1

Since the denominators are positive, the inequality is equivalent to

(n + 1)(n + 2) < $(n + 2)(n + 2)^2 - (n + 1)(n + 2)^2 = (n + 2)^2.$

by multiplying both sides of the inequality with the denominators.

Answer 2

Base case: Assuming $n \ge 1$ and $n \in \mathbb{N}$, I would suggest a proof by induction. If we let $$p(n):=\frac{1}{\left(n+2\right)^2} \lt \frac{1}{n+1} - \frac{1}{n+2}$$ Then $p(1)$ is true, since $$\frac{1}{\left(1+2\right)^2}=\frac{1}{9} \lt \frac{1}{1+1} - \frac{1}{1+2} = \frac{1}{6}$$ Induction hypothesis: Assuming $p(n)$ is true for $n=k$, then $$\frac{1}{\left(k+2\right)^2} \lt \frac{1}{k+1} - \frac{1}{k+2}$$ Setting $n=k+1$, $p(k+1)$ can be found as follows: $$\frac{1}{\left(k+1+2\right)^2}=\frac{1}{\left(k+3\right)^2}=\frac{1}{k^2+6k+9}=\frac{1}{(k^2+5k+6)+k+3}$$ Now $$\frac{1}{(k^2+5k+6)+k+3} \lt \frac{1}{k^2+5k+6}$$ since $k \ge 0$.

Therefore, factoring and using partial fractions: $$\frac{1}{\left(k+3\right)^2} \lt \frac{1}{k^2+5k+6}=\frac{1}{(k+2)(k+3)}= \frac{1}{k+2}-\frac{1}{k+3}$$ Which can be alternatively written as $$\frac{1}{\left[\left(k+1\right) + 2\right]^2} \lt \frac{1}{\left(k+1\right)+1}-\frac{1}{\left(k+1\right) +2}$$ Hence $p(k) \Rightarrow p(k+1)$

Therefore by the principle of mathematical induction $p(n)$ as defined above is true for all $n \in \mathbb{N}$.