MHB Sum Infinity Express: Rational Number Solution

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The discussion focuses on evaluating the double sum $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{m^2n+mn^2+2mn}$ and expressing it as a rational number. The sum is transformed into a more manageable form, leading to the expression $S=\sum_{n=1}^{\infty} \frac{H_{n+2}}{n(n+2)}$, where $H_n$ denotes the harmonic number. Various identities and summation techniques are employed, including the use of partial fractions and telescoping sums, to simplify the terms. Ultimately, the result is determined to be $S = \frac{7}{4}$. The thread showcases multiple approaches to arrive at the same conclusion, emphasizing the collaborative nature of mathematical problem-solving.
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Express $\displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{1}{m^2n+mn^2+2mn}$ as a rational number.
 
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anemone said:
Express $\displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{1}{m^2n+mn^2+2mn}$ as a rational number.

[sp]Writing the sum in aslight different form You have...$\displaystyle S=\sum_{n=1}^{\infty} \frac{1}{n}\ \sum _{m=1}^{\infty} \frac{1}{m\ (m+n+ 2)}\ (1)$In... http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426.html#post2494... it has been demonstrated that is... $\displaystyle\sum_{m=1}^{\infty} \frac{1}{m (m+n)} = \frac{H_{n}}{n}\ (2)$... where $\displaystyle H_{n} = \sum_{k=1}^{n} \frac{1}{k}$ is the harmonic number of order n.In this case is... $\displaystyle S=\sum_{n=1}^{\infty} \frac{H_{n+2}}{n\ (n+2)}\ (3)$

Using the identity $\displaystyle H_{n+2} = H_{n} + \frac{1}{n+1} + \frac{1}{n+2}$ You arrive to write...

$\displaystyle S= \sum_{n=1}^{\infty} \frac{H_{n}}{n\ (n+2)} + \sum_{n=1}^{\infty} \frac{1}{n\ (n+1)\ (n+2)} + \sum_{n=1}^{\infty} \frac{1}{n\ (n+2)^{2}}\ (4)$

The first term of (4) can be found applying the general formula...

$\displaystyle \sum_{n=1}^{\infty} \frac{H_{n}}{n\ (n+a)} = \frac{1}{2\ a}\ \{2\ \zeta(2) + H_{a-1}^{2} + H_{a-1}^{(2)} \}\ (5)$

... where $\displaystyle H_{a-1}^{(2)} = \sum_{k=1}^{a-1} \frac{1}{k^{2}}$. For a=2 is...

$\displaystyle \sum_{n=1}^{\infty} \frac{H_{n}}{n\ (n+2)} = \frac{1+\zeta(2)}{2}\ (6)$

The other terms are more comfortable...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n\ (n+1)\ (n+2)} = \frac{1}{4}\ (7)$

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n\ (n+2)^{2}} = 1 - \frac{\zeta(2)}{2}\ (8)$

... so that the final result is $S= \frac{7}{4}$ [/sp]

Kind regards

$\chi$ $\sigma$
 
Slight variation on chisigma's proof:
[sp]$$\begin{aligned}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{m^2n+mn^2+2mn} &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac1n\frac1{m(m+n+2)} \\ &= \sum_{n=1}^{\infty} \frac1{n(n+2)} \sum_{m=1}^{\infty}\Bigl(\frac1m - \frac1{m+n+2}\Bigr) \qquad\text{(partial fractions)} \\ &= \sum_{n=1}^{\infty} \frac1{n(n+2)} \sum_{m=1}^{n+2}\frac1m \qquad\text{(telescoping sum)} \\ &= \sum_{n=1}^{\infty} \frac12\Bigl(\frac1n - \frac1{n+2}\Bigr) \sum_{m=1}^{n+2}\frac1m \qquad\text{(partial fractions)} \\ &= \frac12\biggl(1 + \frac34 + \sum_{n=1}^{\infty}\frac1n\Bigl(\frac1{n+1} + \frac1{n+2}\Bigr)\biggr) \qquad\text{(partially telescoping sum)*} \\ &= \frac78 + \frac12\sum_{n=1}^{\infty}\Bigl(\frac1n - \frac1{n+1} + \frac12\Bigl(\frac1n - \frac1{n+2}\Bigr)\Bigr) \qquad\text{(partial fractions)} \\ &= \frac78 + \frac12\Bigl(1 + \frac34\Bigr) = \frac74 \qquad\text{(telescoping sums)}.\end{aligned}$$

* The idea of the "partially telescoping sum" is that the $n$th term in the sum $$\sum_{n=1}^{\infty} \Bigl(\frac1n - \frac1{n+2}\Bigr) \sum_{m=1}^{n+2}\frac1m$$ contains the product $$ \frac1n \sum_{m=1}^{n+2}\frac1m$$. Two terms earlier in the sum (assuming that $n\geqslant3$), there is the product $$-\frac1n \sum_{m=1}^{n}\frac1m$$. When those two expressions are added, most of the terms in the "$m$" summations cancel, and we are just left with $$\frac1n\Bigl(\frac1{n+1} + \frac1{n+2}\Bigr).$$[/sp]
 
Hey chisigma and Opalg,(Wave)

Thanks for participating and showing the two neat and smart solutions.

A solution that I saw somewhere online:

$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{1}{m^2n+mn^2+2mn}$

$= \sum_{n=1}^{\infty}\dfrac{1}{n}\sum_{m=1}^{\infty} \dfrac{1}{m(m+n+2)}$

$=\sum_{n=1}^{\infty}\dfrac{1}{n(n+2)}\left[\left(1-\dfrac{1}{n+3} \right)+\left(\dfrac{1}{2}-\dfrac{1}{n+4} \right)+\left(\dfrac{1}{3}-\dfrac{1}{n+5}+\cdots \right) \right]$

$=\sum_{n=1}^{\infty}\dfrac{1}{n(n+2)}\left[1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n+2} \right]$

$=\dfrac{1}{2}\sum_{n=1}^{\infty} \left(\dfrac{1}{n}-\dfrac{1}{n+2} \right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+2} \right)$

$=\dfrac{1}{2}\left[ \left(1-\dfrac{1}{3} \right)\left(1+\dfrac{1}{2}+\dfrac{1}{3} \right)+\left(\dfrac{1}{2} -\dfrac{1}{3} \right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{3} -\dfrac{1}{5} \right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)+\cdots\right]$

$=\dfrac{1}{2}\left[\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)+ \dfrac{1}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\dfrac{1}{3}\left(1+\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{4}\left(1+\dfrac{1}{5}+\dfrac{1}{6}\right)+\dfrac{1}{5}\left(1+\dfrac{1}{6}+\dfrac{1}{7}\right)+\cdots\right]$

$=\dfrac{1}{2}\left[\dfrac{11}{6}+\dfrac{1}{2}\cdot\dfrac{25}{12}+\left(\dfrac{1}{3(4)}+\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\cdots \right)+\left(\dfrac{1}{3(5)}+\dfrac{1}{4(6)}+\dfrac{1}{5(7)}+\cdots \right) \right]$

$=\dfrac{1}{2}\left[\dfrac{11}{6}+\dfrac{25}{24}+\left(\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\cdots\right)+\dfrac{1}{2}\left(\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\left(\dfrac{1}{4}-\dfrac{1}{6}\right)+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\cdots\right)\right]
$

$=\dfrac{1}{2}\left[\dfrac{11}{6}+\dfrac{25}{24}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}\right]$

$=\dfrac{7}{4}$
 
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