Question

`\rm\sum_(n=1)^(\infty)\sum_(m=1)^(\infty) \frac{( - 1 {)}^(n + 1) }{ {mn}^(2) + mn + {m}^(2) n} \\`​

Answer 1

Let
`S` denote the sum. We can first resolve the sum in
`m` by factorizing and decomposing into partial fractions.

`\displaystyle S = \sum_(n=1)^\infty \sum_(m=1)^\infty ((-1)^(n+1))/(mn^2 + mn + m^2n) \\\\ ~~~~ = \sum_(n=1)^\infty \frac{(-1)^(n+1)}n \sum_(m=1)^\infty \frac1{m(m+n+1)} \\\\ ~~~~ = \sum_(n=1)^\infty ((-1)^(n+1))/(n(n+1)) \sum_(m=1)^\infty \left(\frac1m - \frac1{m+n+1}\right)`

Rewrite the
`m`-summand as a definite integral. Interchange the integral and sum, and evaluate the resulting geometric sums.

`\displaystyle \sum_(m=1)^\infty \left(\frac1m - \frac1{m+n+1}\right) = \sum_(m=1)^\infty \int_0^1 \left(x^(m-1) - x^(m+n)\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_(m=1)^\infty \left(x^(m-1) - x^(m+n)\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 (1 - x^(n+1))/(1 - x) \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_(\ell=0)^n x^\ell \, dx \\\\ ~~~~~~~~ = \sum_(\ell=0)^n \int_0^1 x^\ell \, dx \\\\ ~~~~~~~~ = \sum_(\ell=0)^n \frac1{\ell+1} \\\\ ~~~~~~~~ = \sum_(\ell=1)^(n+1) \frac1\ell = H_(n+1)`

where

`H_n = \displaystyle \sum_(\ell=1)^n \frac1\ell = 1 + \frac12 + \frac 13 + \cdots + \frac1n`

is the
`n`-th harmonic number. The generating function will be useful:

`\displaystyle \sum_(n=1)^\infty H_n x^n = -(\ln(1-x))/(1-x)`

To evaluate the remaining sum to get
`S`, let

`\displaystyle f(x) = \sum_(n=1)^\infty (H_(n+1))/(n(n+1)) x^(n+1)`

and observe that
`S=\lim\limilts_(x\to-1^+) f(x)`, which I'll abbreviate to
`f(-1)`. Differentiating twice, we have

`\displaystyle f'(x) = \sum_(n=1)^\infty \frac{H_(n+1)}n x^n`

`\displaystyle f''(x) = \sum_(n=1)^\infty H_(n+1) x^n`

`\displaystyle \implies f''(x) = -(\ln(1-x))/(x^2(1-x)) - \frac1x`

By the fundamental theorem of calculus, noting that
`f(0)=f'(0)=0`, we have

`\displaystyle \int_(-1)^0 f'(x) \, dx = f(0) - f(-1) \implies f(-1) = -\int_(-1)^0 f'(x) \, dx`

`\displaystyle \int_x^0 f''(x) \, dx = f'(0) - f'(x) \implies f'(x) = -\int_x^0 f''(t) \, dt`

`\displaystyle \implies S = f(-1) = \int_(-1)^0 \int_x^0 \left((\ln(1-t))/(t^2(1-t)) + \frac1t\right) \, dt \, dx`

Change the order of the integration, and substitute
`t=-u`.

`S = \displaystyle \int_(-1)^0 \int_(-1)^t \left((\ln(1-t))/(t^2(1-t)) + \frac1t\right) \, dx \, dt \\\\ ~~~ = - \int_(-1)^0 \left(((1+t) \ln(1-t))/(t^2(1-t)) + \frac1t + 1\right) \, dt \\\\ ~~~ = -1 - \int_(-1)^0 \left(\left(\frac2{1-t} + \frac2t + \frac1{t^2}\right) \ln(1-t) + \frac1t\right) \, dt \\\\ ~~~ = -1 - \int_0^1 \left(\left(\frac2{1+u} - \frac2u + \frac1{u^2}\right) \ln(1+u) - \frac1u\right) \, du`

For the remaining integrals, substitute and use power series.

`\displaystyle \int_0^1 (\ln(1+u))/(1+u) \, du = \int_0^1 \ln(1+u) d(\ln(1+u)) = \frac{\ln^2(2)}2`

`\displaystyle \int_0^1 \frac{\ln(1+u)}u \, du = - \int_0^1 \frac1u \sum_(k=1)^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_(k=1)^\infty \frac{(-1)^k}k \int_0^1 u^(k-1) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_(k=1)^\infty ((-1)^k)/(k^2) = (\pi^2)/(12)`

`\displaystyle \int_0^1 (\ln(1+u) - u)/(u^2) \, du = - \int_0^1 \frac1{u^2} \left(\sum_(k=1)^\infty \frac{(-u)^k}k + u\right) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\int_0^1 \frac1{u^2} \sum_(k=2)^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = - \sum_(k=2)^\infty \frac{(-1)^k}k \int_0^1 u^(k-2) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\sum_(k=2)^\infty ((-1)^k)/(k(k-1)) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \sum_(k=1)^\infty ((-1)^k)/(k(k+1)) = 1 - 2\ln(2)`

Tying everything together, we end up with

`S = -1 - \left(2 \cdot \frac{\ln^2(2)}2 - 2 \cdot (\pi^2)/(12) + (1-2\ln(2))\right) \\\\ ~~~ = \boxed{\frac{\pi^2}6 - 2 + 2\ln(2) - \ln^2(2)}`