Solve $(n+1)f(n+1)=(n+2)f(n)+1$

Question

I solved a problem here, but I could not solve the functional equation, $(n+1)f(n+1)=(n+2)f(n)+1$ without induction at that time. So here is a way to solve the problem. I want to know if their is another way to do it, which is shorter than what I show here not involving induction:

Problem:

$f(1)=1$ and $(n+1)f(n+1)=(n+2)f(n)+1$, $n\in \mathbb{N}$. Find $f(n)$.

Solution:

Let $g(n)=\frac{f(n)}{n}$.

Then $$(n+1)^2g(n+1)=n(n+2)g(n)+1\\\Rightarrow (n+1)^2[g(n+1)-1]=n(n+2)[g(n)-1]\\\Rightarrow [g(n+1)-1]=\frac{n(n+2)}{(n+1)^2}[g(n)-1]=\frac{n(n+2)(n-1)(n+1)}{(n+1)^2n^2}[g(n-1)-1]\\=\dots =\frac{n!(n+2)!}{2!(n+1)!(n+1)!}[g(1)-1]=\frac{(n+2)}{2(n+1)}[g(1)-1]=0\\\text{(since as $n\to\infty$, $\frac{n+2}{n+1}\rightarrow 1$)}(*)$$

Hence $g(n)=1\Rightarrow f(n)=n\space\space\space \blacksquare$

Also I need to know if can say $(*)$, as well.

Answer 1

Let $g(n)=\dfrac{f(n)}{n+1}.$ Then $$g(n+1)-g(n)=\dfrac{1}{(n+1)(n+2)}=\dfrac{1}{n+1}-\dfrac{1}{n+2}.$$ Now telescope !

Answer 2

Plugging in $n=1$ into the equation yields $f(2) = 2$. This, along with $f(1)=1$, leads to the irresistible conjecture that $\forall n f(n) = n$. Indeed, we see this is a solution since the LHS becomes $(n+1)^2$ and the RHS becomes $(n+2)n+1 = (n+1)^2$.

Now to prove uniqueness: Suppose there is another solution $g(n)$. There must be at least one number $n$ such that $g(n) \not = n$. Now apply the Extreme Principle: take the smallest $n$ (call it $n_0$) such that $g(n_0) \not = n_0$.

Plug in $n_0-1$ into the functional equation:

$$n_0g(n_0) = (n_0+1)(n_0-1) = n_0^2$$

$$g(n_0) = n_0$$

But this is a contradiction, since we assumed $g(n_0) \not = n$.