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?)
 A: You could try the generating form
$$F(x)=\frac{x^4}{2.3.4}+\frac{x^6}{4.5.6}+...\\
\frac{d^3F}{dx^3}=x+x^3+x^5+...=\frac x{1-x^2}$$
Try to integrate the last expression three times, then take the limit as $x\to1$
A: $$
\begin{align}
\sum_{k=1}^\infty\frac1{2k(2k+1)(2k+2)}
&=\frac12\sum_{k=1}^\infty\left(\frac1{2k(2k+1)}-\frac1{(2k+1)(2k+2)}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac1{2k}-\frac2{2k+1}+\frac1{2k+2}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac2{2k}-\frac2{2k+1}\right)-\lim_{n\to\infty}\frac12\left(\frac12-\frac1{2n+2}\right)\\
&=1-\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}-\frac14\\[3pt]
&=\frac34-\log(2)
\end{align}
$$
A: The $n$-th term is
$$\frac12\left(\frac1{2n}-\frac{2}{2n+1}+\frac1{2n+2}\right).$$
The whole series does not telescope but is
$$\frac12\left(\frac12-\frac23+\frac24-\frac25+\frac26-\frac27+\cdots
\right).$$
This is very similar (not identical) to a well-known series...
A: 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.
