Sum of the Given Series (3)

Relevant wiki: Telescoping Series - Sum

By virtue of $\frac1{n(n+1)(n+2)}=\frac12\left(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right),$ we have $\begin{aligned} S&=\sum_{n=1}^\infty\frac1{n(n+1)(n+2)} \\&=\frac12\sum_{n=1}^\infty\left(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right) \\&=\frac 12\left (\frac 1{1\times2}-\frac 1{2\times3}+\frac1{2\times3}-\frac1{3\times4}+\cdots\right ) \\&=\frac14=\boxed{0.25}. \end{aligned}$

$\begin{aligned} S & = \frac 1{1 \times 2 \times 3} + \frac 1{2 \times 3 \times 4} + \frac 1{3 \times 4 \times 5} + \cdots \\ & = \sum_{k=1}^\infty \frac 1{k(k+1)(k+2)} \\ & = \sum_{k=1}^\infty \frac 12 \left(\frac 1k - \frac 2{k+1} + \frac 1{k+2}\right) \\ & = \frac 12\left(\sum_{k=1}^\infty \left(\frac 1k - \frac 1{k+1}\right) - \sum_{k=1}^\infty \left(\frac 1{k+1} - \frac 1{k+2}\right) \right) \\ & = \frac 12\left(\frac 11 - \frac 12 \right) \\ & = \frac 14 = \boxed{0.25} \end{aligned}$

We have

$\begin{aligned} \dfrac{1}{(n - 1)(n)(n + 1)} &= \dfrac{1}{n(n^2 - 1)} \\ &= \dfrac{n}{n^2 - 1} - \dfrac{1}{n} \\ &= \dfrac{n}{(n - 1)(n + 1)} - \dfrac{1}{n} \\ &= \dfrac{1}{2} \left ( \dfrac{1}{n - 1} + \dfrac{1}{n + 1} \right ) - \dfrac{1}{n} \\ &= \dfrac{1}{2} \left ( \left ( \dfrac{1}{n - 1} - \dfrac{1}{n} \right ) + \left( \dfrac{1}{n + 1} - \dfrac{1}{n} \right ) \right ). \\ \end{aligned}$

Thus,

$\begin{aligned} \sum_{n = 2}^{\infty} \dfrac{1}{(n - 1)(n)(n + 1)} &= \sum_{n = 2}^{\infty} \left ( \dfrac{1}{2} \left ( \left ( \dfrac{1}{n - 1} - \dfrac{1}{n} \right ) + \left( \dfrac{1}{n + 1} - \dfrac{1}{n} \right ) \right ) \right ) \\ &= \dfrac{1}{2} \left (1 - \dfrac{1}{2} \right ) \\ &= \boxed{\dfrac{1}{4}}. \end{aligned}$

For any positive integer n, 1/[(n).(n+1).(n+2)] = (1/2)[1/(n)(n+1) - 1/(n+1)(n+2)]. So 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + ....= (1/2)[1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + 1/(3.4) - 1/(4.5) + ........] = (1/2)[1/(1.2)] = 0.25