<s>[INST] Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper

Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
  and find the sum if it exists.

I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
 [/INST] Let me add a more general answer.
Note that
$$
{1 \over {\left( {x + 1} \right)\left( {x + 2} \right) \cdots \left( {x + m} \right)}}
 = {1 \over {\left( {x + 1} \right)^{\,\overline {\,m\,} } }}
 = {{\Gamma \left( {x + 1} \right)} \over {\Gamma \left( {x + 1 + m} \right)}}
 = x^{\,\underline {\, - m\,} } 
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial.
In virtue of the definition through the Gamma function, the above is valid for $x,m \in \mathbb C$.
Then we apply the Indefinite Sum concept, by which
$$
\eqalign{
  & \Delta _{\,x} \;x^{\,\underline {\,q\,} }
  = \left( {x + 1} \right)^{\,\underline {\,q\,} }  - x^{\,\underline {\,q\,} }
  = qx^{\,\underline {\,q - 1\,} } \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \Delta _{\,x} ^{\left( { - 1} \right)} \;x^{\,\underline {\,q\,} }
  = \sum\nolimits_x {x^{\,\underline {\,q\,} } }  = \left\{ {\matrix{
   {{1 \over {q + 1}}\;x^{\,\underline {\,q + 1\,} }  + c} & { - 1 \ne q}  \cr 
   {\psi (x + 1) + c} & { - 1 = q}  \cr 
 } } \right. \cr} 
$$
which is to recall that it is valid for $x,q,c \in \mathbb C$.
So in particular, for $q=-m$ and $m \ne 1$, the Indefinite sum becomes
$$
\sum\nolimits_x {x^{\,\underline {\, - m\,} } } \quad \left| {\,1 \ne m} \right.
 = {1 \over { - m + 1}}x^{\,\underline {\, - m + 1\,} }  + c
$$
which means that for the definite sum we get
$$
\eqalign{
  & \sum\limits_{x = a}^b {x^{\,\underline {\, - m\,} } }
  = \sum\nolimits_{x = a}^{b + 1} {x^{\,\underline {\, - m\,} } } \quad \left| \matrix{
  \,m,a,b \in C \hfill \cr 
  \;1 \ne m \hfill \cr}  \right. =   \cr 
  &  = {1 \over { - m + 1}}\left( {\left( {b + 1} \right)^{\,\underline {\, - m + 1\,} }
  - a^{\,\underline {\, - m + 1\,} } } \right) =   \cr 
  &  = {1 \over {m - 1}}\left( {a^{\,\underline {\, - \left( {m - 1} \right)\,} }
  - \left( {b + 1} \right)^{\,\underline {\, - \left( {m - 1} \right)\,} } } \right) =   \cr 
  &  = {1 \over {m - 1}}\left( {{1 \over {\left( {a + 1} \right)^{\,\overline {\,m - 1\,} } }}
 - {1 \over {\left( {b + 2} \right)^{\,\overline {\,m - 1\,} } }}} \right) \cr} 
$$
and taking the limit $b \to \infty$
$$
\sum\limits_{x = a}^\infty  {x^{\,\underline {\, - m\,} } } \quad \left| {\,\;1 < m \in R} \right.
 = {1 \over {\left( {m - 1} \right)m!}}
$$
Finally, in your particular case, the summand, the indefinite, definite and infinite sums are
$$
\eqalign{
  & {1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}} = {1 \over {n^{\,\overline {\,3\,} } }}
 = \left( {n - 1} \right)^{\,\underline {\, - 3\,} }   \cr 
  & \sum\nolimits_n {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = \sum\nolimits_n {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } }  =   \cr 
  &  =  - {1 \over 2}\left( {n - 1} \right)^{\,\underline {\, - 2\,} }  + c
 =  - {1 \over {2n^{\,\overline {\,2\,} } }} + c =  - {1 \over {2n\left( {n + 1} \right)}} + c  \cr 
  & \sum\limits_{n = 1}^m {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = \sum\nolimits_{n = 1}^{m + 1} {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } }  =   \cr 
  &  =  - {1 \over 2}\left( {{1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}
 - {1 \over {1 \cdot 2}}} \right) = {1 \over 2}\left( {{1 \over 2}
 - {1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}} \right)  \cr 
  & \sum\limits_{n = 1}^\infty  {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = {1 \over 4} \cr} 
$$
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