let U_n =Σ_(k=0) ^n (1/(k^2 +k+1)) find a equivalent of U


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Question Number 74352 by mathmax by abdo last updated on 22/Nov/19

$l e t U_{n} = \sum_{k = 0}^{n} \frac{1}{k^{2} + k + 1} f i n d a e q u i v a l e n t o f U_{n} (n \to + \infty)$
	Answered by mind is power last updated on 24/Nov/19
	$let f(z)=(π/(z^2 +z+1)).cot(z) we can find the value of the som pols of f are Z∪{j,j^− } Res(f,n∈Z)=(1/(n^2 +n+1)) ∣f(z)∣≤((∣coth(Re(z))∣)/(∣z^2 +z+1∣))→0 ∣z∣→∞ ⇒∫_C_R f(z)dz=0 Res th⇒2iπΣ_(n∈Z) Res(f,n)+2iπRes(f,j,j^− ) Res(f,j)= (π/((j−j^− ))).((cos(πj))/(sin(πj)))=(π/(i(√3))).((cos(−(π/2)+i((√3)/2)))/(sin(−(π/2)+i(√(3/4)))))=−(π/(i(√3))).((sin(0.5i(√3)))/(cos(((i(√3))/2)))) −(π/(√3)).th(((√3)/2)) Res(f,j^− )=(π/(j−j^− )).((cos(((−π)/2)−((i(√3))/2)))/(sin(−(π/2)−i((√3)/2))))=(π/(i(√3))).((−sin(((i(√3))/2)))/(−cos(((i(√3))/2))))=−(π/(√3))th(((√3)/2)) ⇒2iπΣ_(n∈Z) (1/(1+n+n^2 ))+2iπ(−((2π)/(√3)).th(((√3)/2)))=0 ⇒Σ_(n∈Z) (1/(1+n+n^2 ))=((2π)/(√3)).th(((√3)/2)) Σ_(n∈N) (1/(1+n+n^2 ))+Σ_(n∈N^∗ ) (1/(1−n+n^2 ))=Σ_(n∈Z) (1/(1+n+n^2 )) ⇒Σ_(n∈N^∗ ) (1/(1−n+n^2 ))=Σ_(n∈N^∗ ) (1/((n−1)+1+(n−1)^2 ))=Σ_(n∈N) (1/(1+n+n^2 )) ⇒2Σ_(n∈N) (1/(1+n+n^2 ))=((2π)/(√3)).th(((√3)/2))⇒Σ_(n∈N) (1/(1+n+n^2 ))=(π/(√3))th(((√3)/2))$
	$l e t f (z) = \frac{π}{z^{2} + z + 1} . c o t (z)$ $w e c a n f i n d t h e v a l u e o f t h e s o m$ $p o l s o f f a r e Z \cup {j, \bar{j}}$ $R e s (f, n \in Z) = \frac{1}{n^{2} + n + 1}$ $∣ f (z) ∣⩽ \frac{∣ c o t h (R e (z)) ∣}{∣ z^{2} + z + 1 ∣} \to 0 ∣ z ∣\to \infty$ $\Rightarrow \int_{C_{R}} f (z) d z = 0 R e s t h \Rightarrow 2 i π \sum_{n \in Z} R e s (f, n) + 2 i π R e s (f, j, \bar{j})$ $R e s (f, j) =$ $\frac{π}{(j - \bar{j})} . \frac{c o s (π j)}{s i n (π j)} = \frac{π}{i \sqrt{3}} . \frac{c o s (- \frac{π}{2} + i \frac{\sqrt{3}}{2})}{s i n (- \frac{π}{2} + i \sqrt{\frac{3}{4}})} = - \frac{π}{i \sqrt{3}} . \frac{s i n (0.5 i \sqrt{3})}{c o s (\frac{i \sqrt{3}}{2})}$ $- \frac{π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2})$ $R e s (f, \bar{j}) = \frac{π}{j - \bar{j}} . \frac{c o s (\frac{- π}{2} - \frac{i \sqrt{3}}{2})}{s i n (- \frac{π}{2} - i \frac{\sqrt{3}}{2})} = \frac{π}{i \sqrt{3}} . \frac{- s i n (\frac{i \sqrt{3}}{2})}{- c o s (\frac{i \sqrt{3}}{2})} = - \frac{π}{\sqrt{3}} t h (\frac{\sqrt{3}}{2})$ $\Rightarrow 2 i π \sum_{n \in Z} \frac{1}{1 + n + n^{2}} + 2 i π (- \frac{2 π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2})) = 0$ $\Rightarrow \sum_{n \in Z} \frac{1}{1 + n + n^{2}} = \frac{2 π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2})$ $\sum_{n \in N} \frac{1}{1 + n + n^{2}} + \sum_{n \in N^{}} \frac{1}{1 - n + n^{2}} = \sum_{n \in Z} \frac{1}{1 + n + n^{2}}$ $\Rightarrow \sum_{n \in N^{}} \frac{1}{1 - n + n^{2}} = \sum_{n \in N^{*}} \frac{1}{(n - 1) + 1 + {(n - 1)}^{2}} = \sum_{n \in N} \frac{1}{1 + n + n^{2}}$ $\Rightarrow 2 \sum_{n \in N} \frac{1}{1 + n + n^{2}} = \frac{2 π}{\sqrt{3}} . t h (\frac{\sqrt{3}}{2}) \Rightarrow \sum_{n \in N} \frac{1}{1 + n + n^{2}} = \frac{π}{\sqrt{3}} t h (\frac{\sqrt{3}}{2})$
	Commented by abdomathmax last updated on 24/Nov/19

	$t h a n k x s i r .$
	Commented by mind is power last updated on 24/Nov/19

	$y^{'} r e w e l c o m$
	Commented by mathmax by abdo last updated on 24/Nov/19

	$f (z) = \frac{π c o t a n (π z)}{z^{2} + z + 1}$
	Commented by mind is power last updated on 02/Dec/19

	$yes sir$
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