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.
 A: Using Euler's beta function,
$$S=\sum_{k\geq 1}\frac{(k-1)!}{(k+2)!}=\sum_{k\geq 1}\frac{\Gamma(k)}{\Gamma(k+3)}=\frac{1}{2}\sum_{k\geq 1}B(k,3)$$
hence:
$$ S = \frac{1}{2}\int_{0}^{1}(1-x)^2\sum_{k\geq 1}x^{k-1}\,dx=\frac{1}{2}\int_{0}^{1}(1-x)\,dx = \color{red}{\frac{1}{4}}. $$
A straightforward generalization of this approach gives the identity:
$$ \sum_{n\geq 1}\frac{1}{n(n+1)\cdots(n+N)}=\color{red}{\frac{1}{N\cdot N!}}.$$
A: There is an alternate method and is as follows.
Notice that 
$$ \frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3) $$
where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} \\
&= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=1}^{\infty} x^{n-1} \right) \, (1-x)^{2} \, dx \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2}}{1-x} \, dx = \frac{1}{2} \, \int_{0}^{1} (1-x) \, dx \\
&= \frac{1}{4}
\end{align}
This leads to the known result
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} = \frac{1}{4}.
\end{align}
A: Writing 
$$\frac{1}{n(n+1)(n+2)} = \frac{1}{n+1}\left(\frac{1}{2n} - \frac{1}{2(n+2)}\right) = \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)},$$
we see that the sum telescopes to 
$$\frac{1}{2(1)(2)} = \frac{1}{4}.$$
A: Alternatively, take
$$ \frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}, $$
and integrate three times with lower limit $0$, giving
$$ \begin{align*} 
-\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} \\
x + (1-x)\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} \\
\frac{3}{4}x^2 - \frac{1}{2}x - \frac{1}{2} (1-x)^2 \log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)(n+2)}, 
\end{align*}$$
and (as @Clement C reminds me) we then apply Abel's theorem to take the limit as $x \to 1$, which gives $1/4$ as the answer.
A: You are on the right track. Fix $N \geq 1$. Then
$$\begin{align}
\sum_{n=1}^{N} \frac{1}{n \cdot (n+1) \cdot (n+2)} 
&= \sum_{n=1}^N \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}\right)
= \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n+1}+\frac{1}{2}\sum_{n=1}^{N} \frac{1}{n+2}\\
&= \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=2}^{N+1}\frac{1}{n}+\frac{1}{2}\sum_{n=3}^{N+2} \frac{1}{n} \\
&= \frac{1}{2}\left(1+\frac{1}{2}\right) + \frac{1}{2}\sum_{n=3}^N \frac{1}{n}-\left(\frac{1}{2}+\frac{1}{N+1}\right)-\sum_{n=3}^{N}\frac{1}{n}+\\
&\quad \left(\frac{1}{N+1}+\frac{1}{N+2}\right)+\frac{1}{2}\sum_{n=3}^{N} \frac{1}{n} \\
\end{align}$$
Can you continue from there? (the partial sums cancel out, and you only have a few remaining terms. Taking the limit $N\to\infty$ will give you the limit.)
A: Here's another way to do it:$$\sum_{n\ge 1}\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\sum_{n\ge 1}\bigg(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\bigg)=\frac{1}{2}\cdot\frac{1}{1\cdot 2}=\frac{1}{4}.$$One advantage is an easy generalisation to a problem for which the partial fractions decomposition would get thorny:$$\sum_{n\ge 1}\frac{1}{\prod_{j=0}^k(n+j)}=\frac{1}{k!\cdot k}.$$
A: Hint
$$\frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2}$$
Now multiply both sides by $\frac{1}{n+1}$.
A: Using Partial Fraction Decomposition, $$\frac1{n(n+1)(n+2)}=\frac An+\frac B{n+1}+\frac C{n+2}$$
$$\implies 1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$
$$\implies 1=n^2(A+B+C)+n(3A+2B+C)+2A$$
Comparing the coefficients  of the different powers (namely, $0,1,2$) of $n,$ we get $A=\frac12,B=-1,C=\frac12$
$$\implies\frac1{n(n+1)(n+2)}=\frac12\cdot\frac1n-\frac1{n+1}+\frac12\cdot\frac1{n+2}$$
$$=-\frac12\left(\underbrace{\frac1{n+1}-\frac1n}\right)+\frac12\left(\underbrace{\frac1{n+2}-\frac1{n+1}}\right)$$
Can you  recognize the two Telescoping series?
A: Hint. You may write
$$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right)
$$ giving two telescoping sums
$$
\sum_{n=1}^N\frac{1}{n(n+1)(n+2)} =\frac{1}{2}-\frac{1}{2(N+1)}+\frac{1}{2(N+2)}-\frac{1}{4}.
$$
A: 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
$$
\eqalign{
  & {1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}} = {1 \over {n^{\,\overline {\,3\,} } }}
 = \left( {n - 1} \right)^{\,\underline {\, - 3\,} }   \cr 
  & \sum\nolimits_n {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = \sum\nolimits_n {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } }  =   \cr 
  &  =  - {1 \over 2}\left( {n - 1} \right)^{\,\underline {\, - 2\,} }  + c
 =  - {1 \over {2n^{\,\overline {\,2\,} } }} + c =  - {1 \over {2n\left( {n + 1} \right)}} + c  \cr 
  & \sum\limits_{n = 1}^m {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = \sum\nolimits_{n = 1}^{m + 1} {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } }  =   \cr 
  &  =  - {1 \over 2}\left( {{1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}
 - {1 \over {1 \cdot 2}}} \right) = {1 \over 2}\left( {{1 \over 2}
 - {1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}} \right)  \cr 
  & \sum\limits_{n = 1}^\infty  {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = {1 \over 4} \cr} 
$$
