Functional equation problem: $ f ( x ) + \frac 1 { x + 1 } = f ( x + 1 ) $

Question

I've been trying to find a function that satisfies this to solve a separate problem, but I'm finding it difficult and no polynomial seems to work.

$$ f ( x ) + \frac 1 { x + 1 } = f ( x + 1 ) $$

Answer 1

$f(x)+\dfrac{1}{x+1}=f(x+1)$

$f(x+1)-f(x)=\dfrac{1}{x+1}$

$f(x)=\sum\limits_x\dfrac{1}{x+1}+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with unit period

Since $\lim\limits_{x\to+\infty}\dfrac{1}{x+1}=0$, so according to http://en.wikipedia.org/wiki/Indefinite_sum#Mueller.27s_formula, the result can be further simplified to

$f(x)=\sum\limits_{n=0}^\infty\left(\dfrac{1}{n+1}-\dfrac{1}{x+n+1}\right)+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with unit period