Tricky FE

[Nirjhor]

Find all functions $f:\mathbb{N}\mapsto\mathbb{Q}$ satisfying \[f(n+1)=f(n)+\dfrac{1}{n^2+5n+6}\] for all natural numbers $n$.

[mutasimmim]

Fix $f(1)$ any rational number. Then all other $f(n)$ are uniquely determined by the given relation and all of them must also be rational. Thus there are infinitely many solutions.

[Nirjhor]

What are the solutions?

[mutasimmim]

$f(n)=f(1)+ \frac {1}{2.3} +\frac {1}{3.4} + \frac {1}{4.5}+....+\frac{1}{(n+1)(n+2)}= f(1)+\frac{1}{2}-\frac{1}{3}+\frac {1}{3}-\frac{1}{4}+\frac{1}{4}+....+\frac{1}{n+1}-\frac{1}{n+2}$=$ f(1)+\frac{1}{2}-\frac {1}{n+2}$. And $f(1)$ is determined by our choice.



P.S: I don't see any 'tricky' part.

[*Mahi*]

\[f(n+1)=f(n) + \frac 1{n+2} - \frac 1{n+3}\]
Telescope.

[mutasimmim]

This part is a text book problem.

[Nirjhor]

The trick is the equation can be rewritten as \[f(n+1)+\dfrac{1}{(n+1)+2}=f(n)+\dfrac{1}{n+2}\] so \(f(n)+\dfrac{1}{n+2}\) is constant. Letting the constant \(c\in\mathbb{Q}\) and solving leads to \(f(n)=c-\dfrac{1}{n+2}\).

My intention was to show the idea of constantifying which works for a large class of FEs. See the next FE.