Finding the sum of a conditionally convergent double series

Question

I am interested in the double series

$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m+n}mn}{(m+n)^2}.$$

I believe that this series is not absolutely convergent but converges by rows or columns,

$$\sum_{m=1}^{\infty}\left[\sum_{n=1}^{\infty}\frac{(-1)^{m+n}mn}{(m+n)^2}\right]=\sum_{n=1}^{\infty}\left[\sum_{m=1}^{\infty}\frac{(-1)^{m+n}mn}{(m+n)^2}\right]=S.$$

and would like to find both a proof of convergence and the value of $S$ in closed form.

As a first step, I considered the double power series:

$$F(x,y) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m+n}mn}{(m+n)^2}x^my^n$$

which I believe converges absolutely for $|x|,|y|<1$. For $x=y$, I was able to sum diagonally

$$F(x,x) = \sum_{m=2}^{\infty}\sum_{k=1}^{m-1}\frac{(-1)^{m}k(m-k)}{m^2}x^m =\sum_{m=2}^{\infty}\frac{(-1)^{m}x^m}{m^2}\sum_{k=1}^{m-1}k(m-k)=\sum_{m=2}^{\infty}\frac{(-1)^{m}x^m}{6}\Big(m-\frac{1}{m}\Big)=\frac{1}{6}\Big[\log(1+x)-\frac{x}{(1+x)^2}\Big].$$

and find the limit

$$\lim_{x\rightarrow1-}F(x,x) = \frac{1}{6}\Big(\log2-\frac{1}{4}\Big).$$

Note that the diagonal sum of $F(1,1)$ does not converge, but oscillates between $\pm \infty.$

My conjecture is that $S = \frac{1}{6}\Big(\log2-\frac{1}{4}\Big)$ but I have not found a way to directly sum by rows or columns. I also wonder if some extension of Abel's limit theorem can be applied here.

Did · Accepted Answer · 2014-06-01T10:01:59.683

5

To compute the sum $S$, one can start with the identity $$ \frac1{a^2}=\int_0^\infty \mathrm e^{-at}t\mathrm dt, $$ valid for every positive $a$. Using it for each $a=m+n$ allows to get rid of the denominators $(m+n)^2$ and yields $$ S=\int_0^\infty\sum_{m,n}(-1)^{m+n}mnt\mathrm e^{-(m+n)t}t\mathrm dt=\int_0^\infty\left(\sum_n(-1)^nn\mathrm e^{-nt}\right)^2t\mathrm dt. $$ Identifying the inner sum yields $$ S=\int_0^\infty\left(\frac{-\mathrm e^{-t}}{(1+\mathrm e^{-t})^2}\right)^2t\mathrm dt=\int_0^\infty\frac{\mathrm e^{2t}}{(1+\mathrm e^t)^4}t\mathrm dt. $$ Integrating by parts, one gets $$ S=\int_0^\infty\frac{1+3\mathrm e^t}{6(1+\mathrm e^t)^3}\mathrm dt. $$ The change of variable $x=\mathrm e^t$ reduces this to $$ S=\int_1^\infty\frac{1+3x}{6x(1+x)^3}\mathrm dx. $$ Note that $$ \frac{1+3x}{x(1+x)^3}=\frac1x-\frac1{1+x}-\frac1{(1+x)^2}+\frac2{(1+x)^3}, $$ hence $$ 6S=\left.\log\left(\frac{x}{1+x}\right)+\frac1{1+x}-\frac1{(1+x)^2}\right|_1^\infty=\log2-\frac12+\frac14, $$ hence $$ S=\frac16\log2-\frac1{24}. $$

edited Jun 01 '14 at 10:01

answered May 30 '14 at 07:48

Did

279,727

2

Using the first identity requires a justification in changing the order of $\int_0^{\infty}$ and $\sum$. – Sungjin Kim May 30 '14 at 09:12
@i707107 Yes it does. – Did May 30 '14 at 09:27
1

@Did: How can it be justified? Actually you need to change $\sum_m \sum_n \int$ to $\int \sum_m \sum_n$. – Sungjin Kim May 31 '14 at 01:18
@i707107 You already said that and I perfectly understood the first time, thank you. – Did May 31 '14 at 07:10
@Did: (+1) Never mind. I know how to justify it now. Thank you anyway. – Sungjin Kim Jun 01 '14 at 09:48
@i707107 Great. – Did Jun 01 '14 at 09:54
@Did: There is a slight typo in one of the integral expression of $S$. That should be $\int_0^{\infty} te^{-2t} / (1+ e^{-t})^4 dt$. You have the correct one right after that though. – Sungjin Kim Jun 01 '14 at 09:55

score 2 · Answer 2 · answered May 30 '14 at 07:33

It's certainly not absolutely convergent, because e.g. the $m=n$ terms are all $1/4$.

FWIW, Maple's numerical summation (which uses Levin's U transform) agrees with your conjecture:

evalf(Sum(sum(f,n=1..infinity),m=1..infinity));

                     0.07385786343

identify(%);

$$ - \dfrac{1}{24} + \dfrac{1}{6} \ln(2) $$

Finding the sum of a conditionally convergent double series

2 Answers2