<s>[INST] Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}$.
I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?)
 [/INST] Here is an alternative approach that uses a triple integral. 
We begin by noting that
$$\frac{1}{n} = \int_0^1 x^{n - 1} \, dx, \quad \frac{1}{2n + 1} = \int_0^1 y^{2n} \, dy, \quad \frac{1}{n + 1} = \int_0^1 z^n \, dz.$$
The sum can therefore be written as
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{2n(2n + 1)(2n + 2)} &= \frac{1}{4} \sum_{n = 1}^\infty \frac{1}{n(2n + 1)(n + 1)}\\
&= \frac{1}{4} \sum_{n = 1}^\infty \int_0^1 \int_0^1 \int_0^1 x^{n - 1} y^{2n} z^n \, dx dy dz\\
&= \frac{1}{4} \int_0^1 \int_0^1 \int_0^1 \frac{1}{x} \sum_{n = 1}^\infty (xy^2 z)^n \, dx dy dz \tag1\\
&= \frac{1}{4} \int_0^1 \int_0^1 \int_0^1 \frac{1}{x} \cdot \frac{xy^2 z}{1 - xy^2 z} \, dx dy dz \tag2\\
&= \frac{1}{4} \int_0^1 \int_0^1 \int_0^1 \frac{y^2 z}{1 - xy^2 z} \, dx dy dz\\
&= -\frac{1}{4} \int_0^1 \int_0^1 \Big{[} \ln (1 -x y^2 z) \Big{]}_0^1 \, dy dz\\
&= -\frac{1}{4} \int_0^1 \int_0^1 \ln (1 -y^2 z) \, dz dy \tag3 \\
&= -\frac{1}{4} \int_0^1 \left [\frac{(y^2 z - 1)[\ln (1 - y^2 z) - 1]}{y^2} \right ]_0^1 \, dy\\
&= \frac{1}{4} \int_0^1 dy - \frac{1}{4} \int_0^1 \frac{y^2 - 1}{y^2} \ln (1 - y^2) \, dy\\
&= \frac{1}{4} - \frac{1}{4} \left (2 + 2 \int_0^1 \ln (1 - y^2) \, dy \right ) \tag4\\
&= -\frac{1}{4} -\frac{1}{2} \int_0^1 \ln (1 - y^2) \, dy\\
&= -\frac{1}{4} + \int_0^1 \frac{y(1 - y)}{1 - y^2} \, dy \tag5\\
&= -\frac{1}{4} + \int_0^1 \frac{y}{1 + y} \, dy\\
&= -\frac{1}{4} + \int_0^1 \frac{(1 + y) - 1}{1 + y} \, dy\\
&= -\frac{1}{4} +\int_0^1 dy - \int_0^1 \frac{dy}{1 + y}\\
&= -\frac{1}{4} + 1 - \ln (2)\\
&= \frac{3}{4} - \ln (2).
\end{align*}
Explanation
(1) Interchanging the summation with the triple integration.
(2) Summing the series which is geometric.
(3) Interchanging the order of integration.
(4) Integrating by parts.
(5) Integrating by parts.
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