Summing up

heureka knows how to sum up sequences such as this one in a formal manner. I don't!!. However, I can determine the "closed form" of ALL the denominators: (n^3+3n^2+2n). And then I can use sigma to sum them all up: ∑[4/(n^3+3n^2+2n), n, 1, 998]=0.99999799799799......etc. It clearly converges to 1 as n tends to infinity!!.

Need some help in summing up the following sequence:
4/1.2.3 + 4/2.3.4 + 4/3.4.5 +..............+ 4/998.999.1000

Calculate:
\(\begin{array}{|lcll|} \hline \mathbf{ S_n = 4\cdot \left( \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \right) } \\ \mathbf{ S_n = 4\cdot \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} } \\ \hline \end{array} \)

\(\text{Calculate $ S = \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} $} \)

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \dfrac{1}{n(n+d)} = \dfrac{1}{d}\left(\dfrac{1}{n}- \dfrac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\\\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\\\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

We rearrange:

\(\begin{array}{|rcll|} \hline \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{n+1} - \dfrac{1}{n}\times \dfrac{1}{n+2} \\\\ &=& \left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)- \dfrac{1}{2} \times \left(\dfrac{1}{n} -\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n} - \dfrac{1}{n+1} -\dfrac{1}{2n} + \dfrac{1}{2(n+2)} \\\\ \mathbf{\dfrac{1}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} } \\ \hline \end{array}\)

Telescoping series:

\(\begin{array}{|rcll|} \hline S_n &=& \mathbf{\dfrac{1}{2}} &\mathbf{-}& \mathbf{\dfrac{1}{2}} &\color{red}+& \color{red}\dfrac{1}{6} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{4}} &\color{red}-& \color{red}\dfrac{1}{3} &\color{blue}+& \color{blue}\dfrac{1}{8} \\\\ &\color{red}+& \color{red}\dfrac{1}{6} &\color{blue}-& \color{blue}\dfrac{1}{4} &\color{red}+& \color{red}\dfrac{1}{10} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{8} &\color{red}-& \color{red}\dfrac{1}{5} &\color{green}+& \color{green}\dfrac{1}{12} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{2(n-2)} &\color{green}-& \color{green}\dfrac{1}{n-1} &\color{red}+& \color{red}\dfrac{1}{2n} \\\\ &\color{green}+& \color{green}\dfrac{1}{2(n-1)} &\color{red}-& \color{red}\dfrac{1}{n} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+1)}} \\\\ &\color{red}+& \color{red}\dfrac{1}{2n} &\mathbf{-}& \mathbf{\dfrac{1}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+2)}} \\ \hline \end{array}\)

The part of each term cancelling with part of the next two diagonal terms:
Example:

\(\begin{array}{|lcll|} \hline \dfrac{1}{6}-\dfrac{1}{3}+\dfrac{1}{6} = 0 \\\\ \dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{8} = 0 \\\\ \dfrac{1}{10}-\dfrac{1}{5}+\dfrac{1}{10} = 0 \\\\ \ldots \\\\ \dfrac{1}{2n}-\dfrac{1}{n} + \dfrac{1}{2n} = 0 \\ \hline \end{array}\)

So S is, we have all black terms left :

\(\begin{array}{|rcll|} \hline S &=& \dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4} + \dfrac{1}{2(n+1)} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2(n+1)} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{(n+1)(n+2)} \right) \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)} \\\\ &=& \dfrac{1}{4} \left(1- \dfrac{2}{(n+1)(n+2)} \right) \\\\ &=& \dfrac{(n+1)(n+2)-2}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2)+n+2-2}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2)+n}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2+1)}{4(n+1)(n+2)} \\\\ \mathbf{ S }& \mathbf{=} & \mathbf{ \dfrac{n(n+3)}{4(n+1)(n+2)} } \\ \hline \end{array}\)

\(\large{ \mathbf{ S = \sum \limits_{k=1 }^{n} \dfrac 1{k(k+1)(k+2)} = \dfrac{n(n+3)}{4(n+1)(n+2)} } }\)

\(\begin{array}{|lcll|} \hline \mathbf{ S_n} & \mathbf{ =} & \mathbf{ 4\cdot \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} } \\\\ S_n &=& 4\cdot \dfrac{n(n+3)}{4(n+1)(n+2)} \\\\ \mathbf{ S_n} &\mathbf{=} & \mathbf{ \dfrac{n(n+3)}{(n+1)(n+2)} } \quad & | \quad n = 998 \\\\ S_{998} & {=} & \dfrac{998\cdot(998+3)}{(998+1)(998+2)} \\\\ S_{998} & {=} & \dfrac{998\cdot 1001}{999\cdot 1000} \\\\ S_{998} & {=} & \dfrac{499499}{499500} \\\\ S_{998} & {=} & 0.999\overline{997} \\\\ \hline \end{array}\)