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# Help.

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Can someone please find the sum of $$\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots$$

Apr 1, 2020

#1
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∑[ 2 / (n^3 + 3 n^2 + 2 n) , n, 1, ∞] =It converges to 1/2

Here is the "Partial Sum Formula"

sum_(n=1)^m 2/(n^3 + 3 n^2 + 2 n) = (m^2 + 3 m)/(2 (m + 1) (m + 2))

Apr 1, 2020
edited by Guest  Apr 1, 2020
#2
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Algebra
Compute the sum $$\mathbf{\dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \cdots}$$

$$\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\ \begin{array}{|lcll|} \hline s_n = \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\$$

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

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

telescoping series
$$\begin{array}{|rcll|} \hline s_n &=& \mathbf{\dfrac{1}{1}} &\mathbf{-}& \mathbf{\dfrac{2}{2}} &\color{red}+& \color{red}\dfrac{1}{3} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{2}} &\color{red}-& \color{red}\dfrac{2}{3} &\color{blue}+& \color{blue}\dfrac{1}{4} \\\\ &\color{red}+& \color{red}\dfrac{1}{3} &\color{blue}-& \color{blue}\dfrac{2}{4} &\color{red}+& \color{red}\dfrac{1}{5} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{4} &\color{red}-& \color{red}\dfrac{2}{5} &\color{green}+& \color{green}\dfrac{1}{6} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{n-2} &\color{green}-& \color{green}\dfrac{2}{n-1} &\color{red}+& \color{red}\dfrac{1}{n} \\\\ &\color{green}+& \color{green}\dfrac{1}{n-1} &\color{red}-& \color{red}\dfrac{2}{n} &\mathbf{+}& \mathbf{\dfrac{1}{n+1}} \\\\ &\color{red}+& \color{red}\dfrac{1}{n} &\mathbf{-}& \mathbf{\dfrac{2}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{n+2}} \\ \hline \end{array}$$

The part of each term cancelling with part of the next two diagonal terms:
Example:
$$\begin{array}{|lcll|} \hline \frac{1}{3}-\frac{2}{3}+\frac{1}{3} = 0 \\ \frac{1}{4}-\frac{2}{4}+\frac{1}{4} = 0 \\ \frac{1}{5}-\frac{2}{5}+\frac{1}{5} = 0 \\ \ldots \\ \frac{1}{n}-\frac{2}{n} + \frac{1}{n} = 0 \\ \hline \end{array}$$

So s_n is, we have all black terms left :
$$\begin{array}{|rcll|} \hline s_n &=& \dfrac{1}{1}-\dfrac{2}{2}+\dfrac{1}{2} + \dfrac{1}{n+1} - \dfrac{2}{n+1} + \dfrac{1}{n+2} \\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{1}{2} - \dfrac{1}{n+1} + \dfrac{1}{n+2}} \\ \hline \end{array}$$

$$\lim \limits_{n\to \infty} { \dfrac{1}{n+1}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{1}{n+2} } = 0$$

$$\begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{1}{2} - 0 + 0 \\ &=& \dfrac{1}{2} \\ \hline \end{array}$$

$$\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ \dfrac{1}{2} } } \\ \end{array}$$

Apr 2, 2020