what's up folks?
I'm solving the red book of math problems, problem 16 which is to solve the following recurrence relation:
$\sum_{k=1}^n {n \choose k} a(k) = \frac{n}{n+1}$
PS: ${n \choose k} = \frac{n!}{k!(n-k)!}$ (it is not the Legendre symbol).
for every $n \in \mathbb{N}$
I'm wondering if it's possible to use Moebius Inversion formula with
$G(x) = \frac{x}{x+1}$ (one can define it to be zero if $0<x<1$)
and some $F$ I know one can use differential equations theory to solve this one, but I want to try to solve it by this way
spoiler alert: $a(k) = \frac{(-1)^{k+1}}{k+1}$
Hint
It is known that if $f(n) = \sum\limits_{i=0}^{n}\binom{n}{i}g(i)$ for all $0 \le n \le m$, then $g(n) = \sum_{i=0}^{n} (-1)^{i+n}\binom{n}{i} f(i)$ for $0 \le n \le m$.
This sort of inversion is called Pascal's inversion formula
In order to complete the sum to start from $0$ you can assume that $\alpha(0)=0$.
Edit $$\alpha(n)=(-1)^n\frac{3n^2-n}{2}$$