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?
 A: I  would suggest that you try to make what i mention below instead. I think it's more simple ( and it's a general method to find the sum of this type of series, later in this post i give you the general form of it)
The Trick:
 To find the terms of telescoping sum  we can  add in the numerator the following difference:   the greatest term (in your case $(n+x+2)$ ) minus the lowest term $(n+x)$, and to not alter the fraction divide by  that difference (in your case $2$).
To find that sum.
Take $f(n)=\frac{1}{2(n+x)(n+x+1)}$. Apply the trick
$$ \frac{1}{(n+x) (n+x+1)(n+x+2)}= \frac{(n+x+2 )  -(n+x)} {2(n+x)(n+x+1)(n+x+2)} $$
$$=\frac{1}{2(n+x)(n+x+1)}-\frac{1}{2(n+x+1)(n+x+2)}=f(n)-f(n+1)$$
$$=- \left(f(n+1)-f(n) \right) .$$
You can see that $\lim\limits_{n \to \infty} f(n)=0,$
and $f(1)=\frac{1}{2(x+1)(x+2)}$
Then, use the telescopic sum
$$\sum_{n=1}^{\infty}\frac{1}{(n+x) (n+x+1)(n+x+2)}=- \lim_{n \to \infty} [f(n+1)-f(1)]= \frac{1}{2(x+1)(x+2)}.$$

Extra:
Let's see the general case
In general we calculate the following sum (then take $n \to \infty$
$$\sum^{n}_{k=0}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }. $$
The trick (again)
Every time we have this kind of summation we can use the telescoping sum. To find the terms of telescoping sum in  a easy way we can sum add in the numerator the following term: sum $(ak+b+sa)$ the greatest term", subtract thelowest term" $(ak+b)$, and to not alter the fraction divide by $\frac{1}{sa}$ the resultant  difference  of those terms.
$$\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }=\frac{1}{sa}\frac{(ak+b+sa) -(ak+b)}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }$$
$$=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) } -\frac{1}{sa}\frac{1}{(ak+b+a)\ldots (ak+b+sa) }. $$
Take $f(k)=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) }.$
Then the summation is
$$\frac{-1}{sa}\sum^{n}_{k=0} \left(f(k+1) -f(k)\right) $$
by the telescopic summation it's
$$=\frac{-1}{sa}\left(f(n+1) -f(0)\right)=\frac{-1}{sa}\left(\frac{1}{(an+b)\ldots (an+b+(s-1)a) } -\frac{1}{(b)(b+a)\ldots (+b+(s-1)a) }\right). $$
Take $n \to \infty$, then
$$\sum^{\infty}_{k=0}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }=\frac{1}{sa(b)(b+a)\ldots (+b+(s-1)a) }.$$
Example:
To get your case we make $a=1$, $b=x+1$ and $s=2$
$$\sum^{\infty}_{k=0}\frac{1}{(k+x+1)(k+x+2) (k+x+3) }=\frac{1}{2(x+1)(x+2)  }.$$
A: 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 
$$
\frac {1}{2(x+1)(x+2)}
$$
