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.
 A: $$\begin{align}\sum^\infty_{k=1} \frac{1}{k(k+1)(k+2)} & <\sum^\infty_{k=1} \frac{1}{k(k+0)(k+0)}\\ & =\sum^\infty_{k=1} \frac{1}{k^3}\end{align}$$
It converges.
Franko's telescoping sum:
$$\sum^\infty_{k=1} \frac{1}{k(k+1)(k+2)}=\sum^\infty_{k=1}\left(\frac{1/2}{k(k+1)}-\frac{1/2}{(k+1)(k+2)}\right)$$
$$=\frac14$$
A: Observe that
$$\frac{1}{k(k+1)(k+2)}<\frac1{k^3}$$
Then
$$\sum_{k\ge 1} \frac{1}{k(k+1)(k+2)}<\sum_{k\ge 1}\frac1{k^3}$$
I dont know an easy way to calculate it limit. But if you know something about finite calculus we can rewrite your summation as
$$\sum_{k\ge 1} \frac{1}{k(k+1)(k+2)}=\sum_{k\ge 0}k^{\underline {-3}}=\frac{k^{\underline {-2}}}{-2}\Bigg|^{\infty}_{0}=\frac{1}{4}$$
A: 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
$$
\begin{gathered}
  \sum\limits_{k = 1}^\infty  {\frac{1}
{{k\left( {k + 1} \right)\left( {k + 2} \right)}}}  = \sum\limits_{k = 0}^\infty  {\frac{1}
{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}}  = \sum\limits_{k = 0}^\infty  {k^{\,\underline {\, - 3\,} } }  = \mathop {\lim }\limits_{n\; \to \;\infty } \sum\limits_{k = 0}^{n - 1} {k^{\,\underline {\, - 3\,} } }  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1}
{2}\left( {n^{\,\underline {\, - 2\,} }  - 0^{\,\underline {\, - 2\,} } } \right)} \right) =  \hfill \\
   = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1}
{2}\left( {\frac{1}
{{\left( {n + 1} \right)\left( {n + 2} \right)}} - \frac{1}
{{1 \cdot 2}}} \right)} \right) = \frac{1}
{4} \hfill \\ 
\end{gathered} 
$$
A: Yes you compare with $\frac{1}{k^3}$, as follows, 
$$\frac{k^3}{k(k+1)(k+2)}\rightarrow 1$$ so the series converges, because $\sum \frac{1}{k^3}$ converges.
