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Question Number 103879 by abony1303 last updated on 18/Jul/20

Commented by abony1303 last updated on 18/Jul/20

$Please, help$
Commented by myear last updated on 18/Jul/20
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	Answered by Ar Brandon last updated on 18/Jul/20
	$(n/(2n^3 +1))=(n/(((2)^(1/3) n+1)((4)^(1/3) n^2 −(2)^(1/3) n+1))) =(a/((2)^(1/3) n+1))+((bn+c)/((4)^(1/3) n^2 −(2)^(1/3) n+1)) =((a((4)^(1/3) n^2 −(2)^(1/3) n+1)+(bn+c)((2)^(1/3) n+1))/(((2)^(1/3) n+1)((4)^(1/3) n^2 −(2)^(1/3) n+1))) (4)^(1/3) a+(2)^(1/3) b=0 , a+c=0 , −(2)^(1/3) a+b+(2)^(1/3) c=1 ⇒b−2(2)^(1/3) a=1 ⇒ 3(4)^(1/3) a=−(2)^(1/3) ⇒a=((−1)/(3(2)^(1/3) )) , c=(1/(3(2)^(1/3) )) , b=(1/3) ⇒(n/(2n^3 +1))=((−1/3(2)^(1/3) )/((2)^(1/3) n+1))+((n/3+1/3(2)^(1/3) )/((4)^(1/3) n^2 −(2)^(1/3) n+1)) Σ_(n=1) ^∞ (n/(2n^3 +1))=Σ_(n=1) ^∞ {((n/3+1/3(2)^(1/3) )/((4)^(1/3) n^2 −(2)^(1/3) n+1))−((1/3(2)^(1/3) )/((2)^(1/3) n+1))} Any idea to proceed ?$
	$\frac{n}{{2 n}^{3} + 1} = \frac{n}{(\sqrt[3]{2} n + 1) (\sqrt[3]{4} n^{2} - \sqrt[3]{2} n + 1)}$ $= \frac{a}{\sqrt[3]{2} n + 1} + \frac{bn + c}{\sqrt[3]{4} n^{2} - \sqrt[3]{2} n + 1}$ $= \frac{a (\sqrt[3]{4} n^{2} - \sqrt[3]{2} n + 1) + (bn + c) (\sqrt[3]{2} n + 1)}{(\sqrt[3]{2} n + 1) (\sqrt[3]{4} n^{2} - \sqrt[3]{2} n + 1)}$ $\sqrt[3]{4} a + \sqrt[3]{2} b = 0, a + c = 0, - \sqrt[3]{2} a + b + \sqrt[3]{2} c = 1$ $\Rightarrow b - 2 \sqrt[3]{2} a = 1 \Rightarrow 3 \sqrt[3]{4} a = - \sqrt[3]{2} \Rightarrow a = \frac{- 1}{3 \sqrt[3]{2}}, c = \frac{1}{3 \sqrt[3]{2}}, b = \frac{1}{3}$ $\Rightarrow \frac{n}{{2 n}^{3} + 1} = \frac{- 1 / 3 \sqrt[3]{2}}{\sqrt[3]{2} n + 1} + \frac{n / 3 + 1 / 3 \sqrt[3]{2}}{\sqrt[3]{4} n^{2} - \sqrt[3]{2} n + 1}$ $\underset{n = 1}{\sum^{\infty}} \frac{n}{{2 n}^{3} + 1} = \underset{n = 1}{\sum^{\infty}} {\frac{n / 3 + 1 / 3 \sqrt[3]{2}}{\sqrt[3]{4} n^{2} - \sqrt[3]{2} n + 1} - \frac{1 / 3 \sqrt[3]{2}}{\sqrt[3]{2} n + 1}}$ $A n y i d e a t o p r o c e e d ?$
	Answered by ~blr237~ last updated on 19/Jul/20
	$Let state f(z)=(z/(z^3 +a^3 )) g(z)=(π/(tan(πz))) a>0 P(fg)={−a,aw_0 ,aw_1 ,−n,n / n∈N ,w_0 =e^(i(π/3)) .....} let state points A(−in),B(−n+in),C(n+in) D(in) for n≥0 δ_n =ABCDis a rectangle (close way) and f(z)=O((1/(∣z∣^2 ))) (•) The residus theorems allows us to say ∫_δ_n f(z)g(z)dz= 2πi Σ_(x∈P(fg)) ind(δ_n ,x)Res(fg,x)=2πi(Σ_(k=0) ^n f(k) +Res(fg, aw_0 )+Res(fg,aw_1 )] because (•) prove that the first term →0 when n→∞ So S=Σ_(n=0) ^∞ f(n)= −[((aw_0 g(aw_0 ))/(3(aw_0 )^2 )) +((aw_1 g(aw_1 ))/(3(aw_1 )^2 ))] S=−(2/3) Re(((aw_0 g(aw_0 ))/((aw_0 )^2 ))) cause w_(1 ) =w_0 ^− and g(z^− )=g^− (z) S=−((2π)/(3a)) Re((1/( w_0 tan(πaw_0 )))) tan(πw_0 )= −i ((e^(i2aπw_0 ) −1)/(e^(i2πaw_0 ) +1)) and 2πaiw_0 =2iπa((1/2)+i((√3)/2))=−π(√3) +iπa tan(πw_0 )=−i ((e^(iπa) e^(−π(√3)) +1)/(e^(iπa) e^(−π(√3)) −1))=−i((e^(−2π(√3)) −1−2ie^(−π(√3)) sin(πa))/(e^(−2π(√3)) +1−2cos(πa)e^(−π(√3)) )) S=((πe^(π(√3)) )/(6a(e^(−2π(√3)) +1−2e^(−π(√3)) cos(πa)))) Re( (((√3)+i)/(sh(π(√3))+isin(πa)))) S=((πe^(2π(√3)) )/(12a(ch(π(√3))−cos(πa))) ×(((√3) sh(π(√3))−sin(πa))/(sh^2 (π(√3))+sin^2 (πa))) Finally S(a)= Σ_(n=1) ^∞ (n/(n^3 + a^3 )) = ((π(√3)e^(2π(√3)) (sh(π(√3))−sin(πa)))/(12a(ch(π(√3))−cos(πa))(sh^2 (π(√3))+sin^2 (πa)))) a>0 Let deduce your by (1/2) S(^3 (√2)) .$
	$L e t s t a t e f (z) = \frac{z}{z^{3} + a^{3}} g (z) = \frac{π}{t a n (π z)} a > 0$ $P (f g) = {- a, {a w}_{0}, {a w}_{1}, - n, n / n \in N, w_{0} = e^{i \frac{π}{3}} . . . . .}$ $l e t s t a t e p o i n t s A (- i n), B (- n + i n), C (n + i n) D (i n) f o r n ⩾ 0$ $δ_{n} = A B C D i s a r e c t a n g l e (c l o s e w a y) a n d f (z) = O (\frac{1}{∣ z ∣^{2}}) (∙)$ $T h e r e s i d u s t h e o r e m s a l l o w s u s t o s a y$ $\int_{δ_{n}} f (z) g (z) d z = 2 π i \sum_{x \in P (f g)} i n d (δ_{n}, x) R e s (f g, x) = 2 π i (\underset{k = 0}{\sum^{n}} f (k) + R e s (f g, {a w}_{0}) + R e s (f g, {a w}_{1})] b e c a u s e$ $(∙) p r o v e t h a t t h e f i r s t t e r m \to 0 w h e n n \to \infty$ $S o S = \underset{n = 0}{\sum^{\infty}} f (n) = - [\frac{{a w}_{0} g ({a w}_{0})}{3 {({a w}_{0})}^{2}} + \frac{{a w}_{1} g ({a w}_{1})}{3 {({a w}_{1})}^{2}}]$ $S = - \frac{2}{3} R e (\frac{{a w}_{0} g ({a w}_{0})}{{({a w}_{0})}^{2}}) c a u s e w_{1} = {\bar{w}}_{0} a n d g (\bar{z}) = \bar{g} (z)$ $S = - \frac{2 π}{3 a} R e (\frac{1}{w_{0} t a n (π {a w}_{0})})$ $t a n (π w_{0}) = - i \frac{e^{i 2 a π w_{0}} - 1}{e^{i 2 π {a w}_{0}} + 1} a n d 2 π {a i w}_{0} = 2 i π a (\frac{1}{2} + i \frac{\sqrt{3}}{2}) = - π \sqrt{3} + i π a$ $t a n (π w_{0}) = - i \frac{e^{i π a} e^{- π \sqrt{3}} + 1}{e^{i π a} e^{- π \sqrt{3}} - 1} = - i \frac{e^{- 2 π \sqrt{3}} - 1 - 2 {i e}^{- π \sqrt{3}} s i n (π a)}{e^{- 2 π \sqrt{3}} + 1 - 2 c o s (π a) e^{- π \sqrt{3}}}$ $S = \frac{π e^{π \sqrt{3}}}{6 a (e^{- 2 π \sqrt{3}} + 1 - 2 e^{- π \sqrt{3}} c o s (π a))} R e (\frac{\sqrt{3} + i}{s h (π \sqrt{3}) + i s i n (π a)})$ $S = \frac{π e^{2 π \sqrt{3}}}{12 a (c h (π \sqrt{3}) - c o s (π a)} \times \frac{\sqrt{3} s h (π \sqrt{3}) - s i n (π a)}{{s h}^{2} (π \sqrt{3}) + {s i n}^{2} (π a)}$ $F i n a l l y$ $S (a) = \underset{n = 1}{\sum^{\infty}} \frac{n}{n^{3} + a^{3}} = \frac{π \sqrt{3} e^{2 π \sqrt{3}} (s h (π \sqrt{3}) - s i n (π a))}{12 a (c h (π \sqrt{3}) - c o s (π a)) ({s h}^{2} (π \sqrt{3}) + {s i n}^{2} (π a))} a > 0$ $L e t d e d u c e y o u r b y \frac{1}{2} S (^{3} \sqrt{2}) .$
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