<s>[INST] Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum

$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$

I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum. 
Is there any easy way to calculate this?
Please someone help.
 [/INST] Rewrite the sum  $\sum_{i=1}^{\infty }\frac {1}{(2i)(2i+1)(2i+2)}$ as 
$$ \sum_{i=1}^{\infty }\frac {(2i+1)-2i}{(2i)(2i+1)(2i+2)} = \sum_{i=1}^{\infty }\frac {1}{(2i)(2i+2)} -\sum_{i=1}^{\infty }\frac {1}{(2i+1)(2i+2)}  $$
Or using partial fractions 
$$ \frac{1}{4} \sum_{i=1}^{\infty} \left(\frac{1}{i} - \frac{1}{i+1}\right) - \sum_{i=1}^{\infty} \left(\frac{1}{2i+1} - \frac{1}{2i+2}\right)  $$ 
The left sum telescopes to $1$, so we get that $ \frac{1}{4} \sum_{i=1}^{\infty} \left(\frac{1}{i} - \frac{1}{i+1}\right) = \frac14$ 
For the right sum 
$$\sum_{i=1}^{\infty} \left(\frac{1}{2i+1} - \frac{1}{2i+2}\right) = \frac13 - \frac14 + \frac15-\frac16 + \dots $$
We use series expansion for $\ln(1+x)$
$$ \ln(1+x) = \sum_{k=1}^{n} \frac{(-1)^{k-1}x^k}{k} = x-\frac{x^2}{2}+\frac{x^3}{3} - \frac{x^4}{4} + \dots $$ 
Plug in $x=1$, to get that 
$$ \ln 2 = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \dots $$ 
Or 
$$\frac13 - \frac14 + \frac15-\frac16 + \dots = \ln 2 - \frac12 $$
and that's our right sum so the final sum is equal to  
$\frac14 - \ln2 + \frac12 = \frac34-\ln2 $
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