𝚺_(n=1) ^∞ 𝚺_(m=1) ^∞ (((−1)^(n+m) )/(nm(n+m)))


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 184144 by paul2222 last updated on 03/Jan/23

$\underset{n = 1}{\sum^{\infty}} \underset{m = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + m}}{nm (n + m)}$
	Answered by SEKRET last updated on 04/Jan/23

	$\underset{n = 1}{\sum^{\infty}} \frac{1}{n} \cdot \underset{m = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + m}}{m \cdot (n + m)}$ $f {(x)}_{x = - 1} = \underset{m = 1}{\sum^{\infty}} \frac{x^{n + m}}{m \cdot (n + m)}$ $f^{'} (x) = \underset{m = 1}{\sum^{\infty}} \frac{x^{n + m - 1}}{m} = x^{n - 1} \cdot \underset{m = 1}{\sum^{\infty}} \frac{x^{m}}{m}$ $f^{'} (x) = x^{n - 1} \cdot \ln (\frac{1}{1 - x}) C = 0$ $\underset{n = 1}{\sum^{\infty}} {(\frac{- 1}{n} \cdot (\int x^{n - 1} \cdot \ln (1 - x) dx))}_{x = - 1}^{C = 0}$ $\underset{n = 1}{\sum^{\infty}} (- \frac{x^{n} \cdot \ln (1 - x) + B_{x} (n + 1, 0)}{n^{2}})$ $\underset{n = 1}{\sum^{\infty}} (\frac{- {(- 1)}^{n} \cdot \ln (2)}{n^{2}} - \frac{B_{- 1} (n + 1; 0)}{n^{2}}) =$ $= - 1 \cdot \ln (2) \cdot \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} + \underset{n = 1}{\sum^{\infty}} \frac{- B_{- 1} (n + 1, 0)}{n^{2}} =$ $= \frac{π^{2} \cdot \ln (2)}{12} + (\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} Ψ^{0} (\frac{n}{2} + \frac{1}{2})}{2 n^{2}} + \underset{n = 0}{\sum^{\infty}} \frac{- Ψ^{0} (\frac{n}{2} + 1)}{2 n^{2}})$ $= \frac{π^{2} \cdot \ln (2)}{12} + (- 0.28159 - 0.337683)$ $\approx 0.0491823$ $A B D U L A Z I Z A B D U V A L I Y E V$
	Commented by paul2222 last updated on 11/Jan/23


	Answered by witcher3 last updated on 05/Jan/23
	$∫_0 ^1 ((ln^2 (1+x))/x)dx=−li_2 (−x)ln(1+x)+∫_0 ^1 ((li_2 (−x))/(1+x))dx....(E) Li_2 (−x)=ζ(2)−ln(1+x)ln(−x)−li_2 (1+x) =li_2 (−x)ln(1+x)−Li_3 (1+x)+ζ(2)ln(1+x) −∫_0 ^1 ((ln(1+x))/(1+x))ln(−x) =−(1/2)ln^2 (1+x)ln(−x)+(1/2)∫((ln^2 (1+x))/x)dx =−2li_2 (−x)ln(1+x)−2Li_3 (1+x)+2ζ(2)ln(1+x)−ln^2 (1+x)ln(−x) li_2 (−x)=ζ(2)−ln(−x)ln(1+x)−li_2 (1+x) ∫((ln^2 (1+x))/x)dx=−2li_3 (1+x)+2li_2 (1+x)ln(1+x)+ln^2 (1+x)ln(−x)+c Let f(x)=Σ_(n≥1) Σ_(m≥1) (((−1)^(n+m) x^(n+m) )/(nm(n+m))),∀x∈[−1,1] f′(x)=ΣΣ(((−1)^(n+m) )/(nm))x^(n+m−1) =(1/x)ΣΣ(((−x)^n (−x)^m )/(n.m)) =(1/x){Σ(((−x)^n )/n)}{Σ(((−x)^m )/m)} Σ_(n≥1) (((−1)^(n−1) x^n )/n)=ln(1+x) f^′ (x)=((ln^(2() (1+x))/x) f(x)=∫((ln^2 (1+x))/x)dx...(E) apply f(x)=−2Li_3 (1+x)+2Li_2 (1+x)ln(1+x)+ln(−x)ln^2 (1+x)+c f(0)=0 lim_(x→0) {−2Li_3 (1+x)+2Li_2 (1+x)ln(1+x)+ln(−x)ln^2 (1+x)+c}=0 ⇔c−2ζ(3)=0 c=2ζ(3) f(x)=−2Li_3 (1+x)+2Li_2 (1+x)ln(1+x)+ln(−x)ln^2 (1+x)+2ζ(3) our Sum=f(1)$
	$\int_{0}^{1} \frac{{l n}^{2} (1 + x)}{x} d x = - {l i}_{2} (- x) l n (1 + x) + \int_{0}^{1} \frac{{l i}_{2} (- x)}{1 + x} d x . . . . (E)$ ${L i}_{2} (- x) = ζ (2) - l n (1 + x) l n (- x) - {l i}_{2} (1 + x)$ $= {l i}_{2} (- x) l n (1 + x) - {L i}_{3} (1 + x) + ζ (2) l n (1 + x)$ $- \int_{0}^{1} \frac{l n (1 + x)}{1 + x} l n (- x)$ $= - \frac{1}{2} {l n}^{2} (1 + x) l n (- x) + \frac{1}{2} \int \frac{{l n}^{2} (1 + x)}{x} d x$ $= - 2 {l i}_{2} (- x) l n (1 + x) - 2 {L i}_{3} (1 + x) + 2 ζ (2) l n (1 + x) - {l n}^{2} (1 + x) l n (- x)$ ${l i}_{2} (- x) = ζ (2) - l n (- x) l n (1 + x) - {l i}_{2} (1 + x)$ $\int \frac{{l n}^{2} (1 + x)}{x} d x = - 2 {l i}_{3} (1 + x) + 2 {l i}_{2} (1 + x) l n (1 + x) + {l n}^{2} (1 + x) l n (- x) + c$ $L e t f (x) = \sum_{n ⩾ 1} \sum_{m ⩾ 1} \frac{{(- 1)}^{n + m} x^{n + m}}{n m (n + m)}, \forall x \in [- 1, 1]$ $f^{'} (x) = Σ Σ \frac{{(- 1)}^{n + m}}{n m} x^{n + m - 1} = \frac{1}{x} Σ Σ \frac{{(- x)}^{n} {(- x)}^{m}}{n . m}$ $= \frac{1}{x} {Σ \frac{{(- x)}^{n}}{n}} {Σ \frac{{(- x)}^{m}}{m}}$ $\sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1} x^{n}}{n} = l n (1 + x)$ $f^{^{'}} (x) = \frac{{l n}^{2 (} (1 + x)}{x}$ $f (x) = \int \frac{{l n}^{2} (1 + x)}{x} d x . . . (E) a p p l y$ $f (x) = - 2 {L i}_{3} (1 + x) + 2 {L i}_{2} (1 + x) l n (1 + x) + l n (- x) {l n}^{2} (1 + x) + c$ $f (0) = 0$ $\lim_{x \to 0} {- 2 {L i}_{3} (1 + x) + 2 {L i}_{2} (1 + x) l n (1 + x) + l n (- x) {l n}^{2} (1 + x) + c} = 0$ $\Leftrightarrow c - 2 ζ (3) = 0$ $c = 2 ζ (3)$ $f (x) = - 2 {L i}_{3} (1 + x) + 2 {L i}_{2} (1 + x) l n (1 + x) + l n (- x) {l n}^{2} (1 + x) + 2 ζ (3)$ $o u r S u m = f (1)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum