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 $

Question

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?)

Answer 1

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...

Answer 2

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 $x\to1$

Answer 3

$$ \begin{align} \sum_{k=1}^\infty\frac1{2k(2k+1)(2k+2)} &=\frac12\sum_{k=1}^\infty\left(\frac1{2k(2k+1)}-\frac1{(2k+1)(2k+2)}\right)\\ &=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac1{2k}-\frac2{2k+1}+\frac1{2k+2}\right)\\ &=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac2{2k}-\frac2{2k+1}\right)-\lim_{n\to\infty}\frac12\left(\frac12-\frac1{2n+2}\right)\\ &=1-\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}-\frac14\\[3pt] &=\frac34-\log(2) \end{align} $$

Answer 4

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.