Showing that $\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$

Question

Prove that if $x>-1$ then $$\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$$

Could I have a hint for this? I tried writing the RHS as a power series and coming up with a differential equation but it doesn't really help as I end up with an intractable integral.

Answer 1

Hint: This is equivalent to

$$\frac{1}{x} = \frac{0!}{(x+1)}+\frac{1!}{(x+1)(x+2)}+\frac{2!}{(x+1)(x+2)(x+3)}+\cdots.$$

Let $$f_n(x)=\frac{n!}{x(x+1)\cdots(x+n)}.$$

We have that

$$f_n(x)=f_{n-1}(x)-f_{n-1}(x+1),$$

and we wish to prove that

$$\frac{1}{x}=\sum_{n=0}^{\infty} f_n(x+1).$$

Can you take it from here?

Answer 2

Another way. from the properties of the Pochhammer symbol and Beta function we can note that $$\sum_{n\geq0}\frac{n!}{\left(x+1\right)\left(x+2\right)\cdots\left(x+n+1\right)}=\sum_{n\geq0}\frac{n!}{\left(x+1\right)_{n+1}}=\sum_{n\geq0}\frac{\Gamma\left(n+1\right)\Gamma\left(x+1\right)}{\Gamma\left(x+n+2\right)}$$ $$=\sum_{n\geq0}\int_{0}^{1}u^{n}\left(1-u\right)^{x}du=\int_{0}^{1}\left(1-u\right)^{x-1}du=\frac{1}{x}.$$