Σ_(m = 1) ^∞ Σ_(n = 1) ^∞ (1/(mn(m+n))) ?


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Question Number 99005 by bobhans last updated on 18/Jun/20

$\underset{m = 1}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{1}{mn (m + n)} ?$
	Answered by maths mind last updated on 18/Jun/20

	$\sum_{m ⩾ 1} . \frac{1}{m^{2}} \sum_{n ⩾ 1} . \frac{(n + m) - n}{n (n + m)}$ $= \sum_{m ⩾ 1} \frac{1}{m^{2}} . \sum_{n ⩾ 1} (\frac{1}{n} - \frac{1}{n + m})$ $= \sum_{m ⩾ 1} \frac{1}{m^{2}} . H_{m}$ $= \sum_{m ⩾ 1} \frac{H_{m}}{m^{2}}$ $\int_{0}^{1} x^{n - 1} l n (1 - x) d x = - \frac{H_{n}}{n}$ $\Rightarrow \sum_{m ⩾ 1} \frac{H_{m}}{m^{2}} = - \int_{0}^{1} \sum_{m ⩾ 1} \frac{x^{m - 1}}{m} l n (1 - x)$ $= \int_{0}^{1} - \frac{1}{x} (\sum_{m ⩾ 1} \frac{x^{m}}{m}) l n (1 - x) d x$ $= \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x} d x$ $= \int_{0}^{1} \frac{{l n}^{2} (t)}{1 - t} d t, l n (t) = - r \Rightarrow$ $= \int_{0}^{\infty} \frac{t^{2} e^{- t} d t}{1 - e^{- t}} = \int_{0}^{+ \infty} \frac{t^{2} d t}{e^{t} - 1} = Γ (3) ζ (3)$ $i u s e d \int_{0}^{+ \infty} \frac{t^{a - 1}}{e^{t} - 1} = Γ (a) ζ (a) f o r R e (a) > 1$ $w e g e t Γ (3) ζ (3) = 2 ζ (3) = \sum_{m ⩾ 1} . \sum_{n ⩾ 1} \frac{1}{m n (n + m)}$
	Answered by bemath last updated on 18/Jun/20
	$∫_0 ^1 ∫_0 ^1 ∫_0 ^1 ((z dxdydz)/((1−xz)(1−yz))) = ∫_0 ^1 z {∫_0 ^1 (dx/(1−xz)) ∫_0 ^1 (dy/(1−yz)) } dz = ∫_0 ^1 z {[ ((−ln(1−xz))/z) ]_0 ^1 × [((−ln(1−yz))/z)]_0 ^1 } dz = ∫_0 ^1 ((( ln(1−z))^2 )/z) dz , set 1−z =t = ∫_0 ^1 ((( ln(t))^2 )/(1−t)) dt = ∫_0 ^1 (ln (t))^2 Σ_(n=0) ^∞ t^n = Σ_(n=0) ^∞ ∫_0 ^1 t^n (ln (t))^2 dt = 2 Σ_(n=0) ^∞ (1/((n+1)^3 )) = 2 Σ_(n=0) ^∞ (1/n^3 ) = 2 ζ(3) ■$
	$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{z dxdydz}{(1 - xz) (1 - yz)} =$ $\int_{0}^{1} z {\int_{0}^{1} \frac{dx}{1 - xz} \int_{0}^{1} \frac{dy}{1 - yz}} dz =$ $\int_{0}^{1} z {{[\frac{- \ln (1 - xz)}{z}]}_{0}^{1} \times {[\frac{- \ln (1 - yz)}{z}]}_{0}^{1}} dz$ $= \int_{0}^{1} \frac{{(\ln (1 - z))}^{2}}{z} dz, set 1 - z = t$ $= \int_{0}^{1} \frac{{(\ln (t))}^{2}}{1 - t} dt = \int_{0}^{1} {(\ln (t))}^{2} \underset{n = 0}{\sum^{\infty}} t^{n}$ $= \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} t^{n} {(\ln (t))}^{2} dt$ $= 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}} = 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{n^{3}}$ $= 2 ζ (3) ◼$
	Commented by maths mind last updated on 18/Jun/20

	$n i c e$ $\frac{1}{(n + 1) . (m + 1) (n + m + 2)}$ $j u s t a l i t t e l j u s t i f a c t i o n$ $\int_{0}^{1} x^{n} d x . \int_{0}^{1} y^{m} d y . \int_{0}^{1} z^{n + m + 1} d z$ $= \frac{1}{n + 1} . \frac{1}{m + 1} . \frac{1}{(n + m + 2)}$ $= \int_{0}^{1} z \int_{0}^{1} {(x z)}^{n} d x . \int_{0}^{1} {(z y)}^{m} d y . d z$ $\sum_{m} . \sum_{n} \int_{0}^{1} z \int_{0}^{1} {(x z)}^{n} d x . \int_{0}^{1} {(y z)}^{m} d y . d z$ $= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{z d x d y d z}{(1 - x z) (1 - y z)} = Σ Σ \frac{1}{n m (n + m)}$
	Commented by bemath last updated on 18/Jun/20

	$great sir$
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