Sum Infinity Express: Rational Number Solution

[anemone]

Express $\displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{1}{m^2n+mn^2+2mn}$ as a rational number.

 

[chisigma]

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

$\chi$ $\sigma$

 

[Opalg]

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

 

[anemone]

Hey chisigma and Opalg,(Wave)

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

A solution that I saw somewhere online:

Spoiler

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

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

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

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

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

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

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

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

$=\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]
$

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

$=\dfrac{7}{4}$

 

[CellCoree]

I would like to clarify that the given expression is not a sum of infinite terms, but rather a double sum over two variables, n and m. To obtain a rational number solution, we can use the fact that the given expression can be rewritten as:

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

We can then use the formula for the sum of a geometric series to write:

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

The sum $\sum_{n=1}^{\infty} \dfrac{1}{n}$ is a well-known infinite series that converges to the natural logarithm of 2 (ln 2). The sum $\sum_{m=1}^{\infty} \dfrac{1}{m(m+n+2)}$ can be evaluated using partial fractions to obtain:

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

Substituting this back into our original expression, we get:

$\displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{1}{mn(m+n+2)} = \ln 2 \cdot \left(\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{5} + ...\right)$

This is an alternating series that converges to ln 2/2. Therefore, the rational number solution to the given expression is simply ln 2/2.