A chain of triplets

$\begin{aligned} S =\dfrac { 1 }{ 1\cdot 2\cdot 3 } +\dfrac { 1 }{ 3\cdot 4\cdot 5 } +\dfrac { 1 }{ 5\cdot 6\cdot 7 } +\dfrac { 1 }{ 7\cdot 8\cdot 9 } +\frac { 1 }{ 9\cdot 10\cdot 11 } +\cdots \\ S =\dfrac { 3-2}{ 1\cdot 2\cdot 3 } +\dfrac { 5-4 }{ 3\cdot 4\cdot 5 } +\dfrac { 7-6 }{ 5\cdot 6\cdot 7 } +\dfrac { 9-8 }{ 7\cdot 8\cdot 9 } +\frac { 11-10 }{ 9\cdot 10\cdot 11 } +\cdots \\ S = \dfrac{1}{2\cdot 3} + \dfrac{1}{3\cdot 4} + \dfrac{1}{5\cdot 6} + \cdots - \left(\dfrac{1}{3\cdot 5} +\dfrac{1}{5\cdot 7} + \dfrac{1}{7\cdot 9} + \cdots\right) \\ S = \dfrac{1}{2} -{\color{#E81990}\dfrac{1}{3}} + \dfrac{1}{3} -\dfrac{1}{4} + \cdots - \dfrac{1}{2}\left(\dfrac{1}{3} -{\color{#3D99F6}\dfrac{1}{5}}+ {\color{#3D99F6}\dfrac{1}{5}} -{\color{#D61F06}\dfrac{1}{7}} + {\color{#D61F06}\dfrac{1}{7}} + \cdots\right) \\ S =1 -1 + \dfrac{1}{2} + \dfrac{1}{3} -\dfrac{1}{4} + \cdots -{\color{#E81990}\dfrac{1}{3}}- \dfrac{1}{2}\times\dfrac{1}{3} \\ S = \ln2 - \dfrac{1}{3}\left(1+\frac{1}{2}\right) = \ln2 - \dfrac{1}{2} \end{aligned}$

Hence, $\ln A-\dfrac{1}{A} = \ln2 -\dfrac{1}{2} \implies A = 2$ .

Use the identity $\frac1{(2n-1)2n(2n+1)} = \frac{1/2}{2n-1} - \frac1{2n} + \frac{1/2}{2n+1}$ to rewrite as $\begin{aligned} \sum_{n=1}^\infty \frac1{(2n-1)2n(2n+1)} &= \left( \frac{1/2}1 - \frac12 + \frac{1/2}3 \right) + \left( \frac{1/2}3 - \frac14 + \frac{1/2}5 \right) + \left( \frac{1/2}5 - \frac16 + \frac{1/2}7 \right) + \cdots \\ &= \frac{1/2}1 - \frac12 + \frac13 - \frac14 + \frac15 - \cdots \\ &= -\frac12 + \left( 1 - \frac12 + \frac13 - \frac14 + \frac15 - \cdots \right) \\ &= -\frac12 + \ln 2. \end{aligned}$ So the answer is $\fbox{2}.$

$\begin{aligned} S & = {\color{#3D99F6}\frac 1{1\cdot 2 \cdot 3}} + {\color{#D61F06}\frac 1{3\cdot 4 \cdot 5}} + {\color{#3D99F6}\frac 1{5\cdot 6 \cdot 7}} + \cdots & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \frac 12 \left({\color{#3D99F6}\frac 11 - \frac 22 + \frac 13} + {\color{#D61F06}\frac 13 - \frac 24 + \frac 15} + {\color{#3D99F6}\frac 15 - \frac 26 + \frac 17} + \cdots \right) \\ & = \frac 12 \left({\color{#D61F06} \frac 11} + \frac 11 - \frac 22 + \frac 23 - \frac 24 + \frac 25 - \frac 26 + \frac 27 - \cdots \right) \color{#D61F06} - \frac 12 \\ & = \left(\frac 11 - \frac 12 + \frac 13 - \frac 14 + \frac 15 - \frac 16 + \frac 17 - \cdots \right) - \frac 12 \\ & = \ln 2 - \frac 12 \end{aligned}$

Therefore, $A = \boxed{2}$ .

Looking at the first term I noticed that $S$ is slightly more than $\frac16$ . I mentally estimated that the value of $\ln A-\frac1A$ would be sufficiently more than $\frac16$ if $A=3$ since $\ln3-\frac13>1-\frac13=\frac23$ . That meant I could confidently guess that the only prime left to be "a little more than $\frac16$ " would have to be $A=2$ .

Not as elegant as some other solutions, but streightforward:

$\begin{aligned} S &= \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots \\ &= \sum_{n = 0}^\infty \frac{1}{(2n + 1)(2n + 2)(2n + 3)} \\ &= \sum_{n = 0}^\infty \left[ \frac{1}{2(2n + 1)} - \frac{1}{2n + 2} + \frac{1}{2(2n + 3)} \right] . \end{aligned}$

The last line can be obtained by partial fraction decomposition. By an index shift the last term can be rewritten in the form of the first,

$\begin{aligned} \sum_{n = 0}^\infty \frac{1}{2(2n + 3)} &= \sum_{n = -1}^\infty \frac{1}{2(2n + 1)} - \frac{1}{2(-2 + 1)} = \sum_{n = 0}^\infty \frac{1}{2(2n + 1)} - \frac{1}{2} . \end{aligned}$

Therefore

$\begin{aligned} S &= \sum_{n = 0}^\infty \Bigg[ \underbrace{\frac{1}{2n + 1}}_{\frac{1}{1} + \frac{1}{3} + \ldots} \underbrace{- \frac{1}{2n + 2}}_{-\frac{1}{2} - \frac{1}{4} - \ldots} \Bigg] - \frac{1}{2} . \end{aligned}$

The denominator of the first fraction runs over all positive odd numbers, while the denominator of the second fraction runs over all positive even numbers. Since the sign is different, we find an alternating harmonic series,

$\begin{aligned} S &= \underbrace{\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n}}_{\ln{2}} - \frac{1}{2} . \end{aligned}$

So $A = 2$ .

I guessed. :D

Plese split the the terms into some partial fractions and then proceed