Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit
I tried expanding out the denominator, and then using the comparison test with $\frac{1}{k^3}$ but I think this is an incorrect use of the comparison test as I get divergence. I know that the limit is $\frac{1}{4}$ but I am not sure how to use the comparison test to then apply the limit.

Given that the rising and falling factorials are defined  as
$$
\begin{array}{l}
 n^{\,\overline {\,m\,} }  = n\left( {n + 1} \right)\, \cdots \;\left( {n + m - 1} \right) \\ 
 n^{\,\underline {\,m\,} }  = n\left( {n - 1} \right)\, \cdots \;\left( {n - \left( {m + 1} \right)} \right) \\ 
 n^{\,\underline {\, - \,m\,} }  = \frac{1}{{\left( {n + m} \right)^{\,\underline {\,m\,} } }} = \frac{1}{{\left( {n + 1} \right)^{\,\overline {\,m\,} } }} \\ 
 \end{array}
$$
where here we consider $n$ and $m$ integers,
and that it is not difficult to demonstrate that
$$
\begin{gathered}
  \Delta _{\,n} \,n^{\,\underline {\,m\,} }  = \left( {n + 1} \right)^{\,\underline {\,m\,} }  - n^{\,\underline {\,m\,} }  = m\,n^{\,\underline {\,m - 1\,} }  \hfill \\
  \sum\limits_{k = a}^{b - 1} {k^{\,\underline {\,m\,} } } \quad \left| {\;a < b} \right.\quad  = \frac{1}
{{m + 1}}\left( {b^{\,\underline {\,m + 1\,} }  - a^{\,\underline {\,m + 1\,} } } \right) \hfill \\ 
\end{gathered} 
$$
Then
$$
\begin{gathered}
  \sum\limits_{k = 1}^\infty  {\frac{1}
{{k\left( {k + 1} \right)\left( {k + 2} \right)}}}  = \sum\limits_{k = 0}^\infty  {\frac{1}
{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}}  = \sum\limits_{k = 0}^\infty  {k^{\,\underline {\, - 3\,} } }  = \mathop {\lim }\limits_{n\; \to \;\infty } \sum\limits_{k = 0}^{n - 1} {k^{\,\underline {\, - 3\,} } }  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1}
{2}\left( {n^{\,\underline {\, - 2\,} }  - 0^{\,\underline {\, - 2\,} } } \right)} \right) =  \hfill \\
   = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1}
{2}\left( {\frac{1}
{{\left( {n + 1} \right)\left( {n + 2} \right)}} - \frac{1}
{{1 \cdot 2}}} \right)} \right) = \frac{1}
{4} \hfill \\ 
\end{gathered} 
$$
