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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?)

CiaPan
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Anu
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4 Answers4

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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 $x\to1$

Empy2
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  • $\displaystyle \int \frac{xdx}{1-x^2}=-\frac{1}{2}log(1-x^2)$. Now I think integrating the function $log(1-x^2)$ is not so easy. – Mini_me Jan 16 '19 at 09:54
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$$ \begin{align} \sum_{k=1}^\infty\frac1{2k(2k+1)(2k+2)} &=\frac12\sum_{k=1}^\infty\left(\frac1{2k(2k+1)}-\frac1{(2k+1)(2k+2)}\right)\\ &=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac1{2k}-\frac2{2k+1}+\frac1{2k+2}\right)\\ &=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac2{2k}-\frac2{2k+1}\right)-\lim_{n\to\infty}\frac12\left(\frac12-\frac1{2n+2}\right)\\ &=1-\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}-\frac14\\[3pt] &=\frac34-\log(2) \end{align} $$

robjohn
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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...

Angina Seng
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  • 1
    Thank you! This worked for me.
    \begin{align} \sum_{k=1}^n \frac{1}{2n(2n+1)(2n+2)} &= \frac{1}{2} \sum_{k=1}^n \left(\frac{1}{2k} - \frac{2}{2k+1} + \frac{1}{2k+2}\right) \ &= \frac{1}{2} \left( \sum_{k=1}^{n} \left(\frac{2}{2k} - \frac{2}{2k+1}\right) - \frac{1}{2} + \frac{1}{2n+2} \right)\ \end{align} Taking limits as $n \rightarrow \infty$, this is $$ \frac{1}{2} \left(- 2\log 2 + 2 - \frac{1}{2} \right) = \frac{3}{4} - \log 2$$     – Anu Jan 26 '18 at 08:08
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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.

omegadot
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