Prove that the series $\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}$ has a sum of $\frac{1}{2(x+1)(x+2)}$ I am trying to prove that the following series:
$$
\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}
$$
has a sum of:
$$
\frac{1}{2(x+1)(x+2)}
$$
What I've tried:
Using the criteria that $S_n = b_1 - l$:
$$
\frac{1}{(n+x)} - \frac{1}{(n+x+1)(n+x+2)} = \frac{(n+x+1)(n+x+2)-(n+x)}{(n+x)(n+x+1)(n+x+2)} \\
\implies \frac{1}{(n+x)(n+x+1)(n+x+2)} = \frac{1}{(n+x+1)(n+x+2)-(n+x)}(\frac{1}{(n+x)} - \frac{1}{(n+x+1)(n+x+2)}) = \frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2} - \frac{1}{(n+x+1)^2(n+x+2)^2-(n+x)(n+x+1)(n+x+2)} = b_n - b_{n+1}
$$
Now that I have:
$$
b_n = \frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2} \\
l = \lim_{n \to \infty}{b_n} = \lim_{n \to \infty}{\frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2}} = \frac{1}{\infty} = 0 \\
b_1 = \frac{1}{(1+x)(1+x+1)(1+x+2) - (1+x)^2} = \frac{1}{(1+x)(2+x)(3+x) -
 (1+x)^2}
$$
Therefore:
$$
S_n = b_1 - l \implies \\
S_n = \frac{1}{(1+x)(2+x)(3+x) -
 (1+x)^2} - 0 \\
S_n = \frac{1}{(1+x)(2+x)(3+x) -
 (1+x)^2} \\
= \frac{1}{(1+x)}.\frac{1}{(2+x)(3+x) -
 (1+x)} \\
= \frac{1}{(1+x)}.\frac{1}{(6+3x+2x+x^2-1-x) -
 (1+x)} \\
= \frac{1}{(1+x)}.\frac{1}{(x^2+4x+5)}
$$
Which is not the expected result. Am I mistaken somewhere or is this completely wrong?

The given expression can be simplified as 
$$\frac {1}{2} \left[ \frac {1}{(n+x)(n+x+1)} - \frac {1}{(n+x+1)(n+x+2)}\right ]$$
Which simply telescopes to 
$$
\frac {1}{2(x+1)(x+2)}
$$
