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

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

$${let}\:{U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\:\:{find}\:{a}\:{equivalent}\:{of}\:{U}_{{n}} \:\:\:\left({n}\rightarrow+\infty\right) \\ $$$$ \\ $$

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))

$${let}\:{f}\left({z}\right)=\frac{\pi}{{z}^{\mathrm{2}} +{z}+\mathrm{1}}.{cot}\left({z}\right) \\ $$$${we}\:{can}\:{find}\:{the}\:{value}\:{of}\:{the}\:{som} \\ $$$${pols}\:{of}\:{f}\:{are}\:\mathbb{Z}\cup\left\{{j},\overset{−} {{j}}\right\} \\ $$$${Res}\left({f},{n}\in\mathbb{Z}\right)=\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$$$\mid{f}\left({z}\right)\mid\leqslant\frac{\mid{coth}\left({Re}\left({z}\right)\right)\mid}{\mid{z}^{\mathrm{2}} +{z}+\mathrm{1}\mid}\rightarrow\mathrm{0}\:\:\:\mid{z}\mid\rightarrow\infty \\ $$$$\Rightarrow\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{0}\:\:\:\:{Res}\:{th}\Rightarrow\mathrm{2}{i}\pi\underset{{n}\in\mathbb{Z}} {\sum}{Res}\left({f},{n}\right)+\mathrm{2}{i}\pi{Res}\left({f},{j},\overset{−} {{j}}\right) \\ $$$${Res}\left({f},{j}\right)= \\ $$$$\frac{\pi}{\left({j}−\overset{−} {{j}}\right)}.\frac{{cos}\left(\pi{j}\right)}{{sin}\left(\pi{j}\right)}=\frac{\pi}{{i}\sqrt{\mathrm{3}}}.\frac{{cos}\left(−\frac{\pi}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{{sin}\left(−\frac{\pi}{\mathrm{2}}+{i}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}\right)}=−\frac{\pi}{{i}\sqrt{\mathrm{3}}}.\frac{{sin}\left(\mathrm{0}.\mathrm{5}{i}\sqrt{\mathrm{3}}\right)}{{cos}\left(\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \\ $$$$−\frac{\pi}{\sqrt{\mathrm{3}}}.{th}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${Res}\left({f},\overset{−} {{j}}\right)=\frac{\pi}{{j}−\overset{−} {{j}}}.\frac{{cos}\left(\frac{−\pi}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{{sin}\left(−\frac{\pi}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}=\frac{\pi}{{i}\sqrt{\mathrm{3}}}.\frac{−{sin}\left(\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{−{cos}\left(\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}=−\frac{\pi}{\sqrt{\mathrm{3}}}{th}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{2}{i}\pi\underset{{n}\in\mathbb{Z}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} }+\mathrm{2}{i}\pi\left(−\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}.{th}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\in\mathbb{Z}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} }=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}.{th}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\underset{{n}\in\mathbb{N}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} }+\underset{{n}\in\mathbb{N}^{\ast} } {\sum}\frac{\mathrm{1}}{\mathrm{1}−{n}+{n}^{\mathrm{2}} }=\underset{{n}\in\mathbb{Z}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} }\: \\ $$$$\Rightarrow\underset{{n}\in\mathbb{N}^{\ast} } {\sum}\frac{\mathrm{1}}{\mathrm{1}−{n}+{n}^{\mathrm{2}} }=\underset{{n}\in\mathbb{N}^{\ast} } {\sum}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)+\mathrm{1}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}\in\mathbb{N}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}\underset{{n}\in\mathbb{N}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} }=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}.{th}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\Rightarrow\underset{{n}\in\mathbb{N}} {\sum}\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} }=\frac{\pi}{\sqrt{\mathrm{3}}}{th}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by abdomathmax last updated on 24/Nov/19

thankx sir.

$${thankx}\:{sir}. \\ $$

Commented by mind is power last updated on 24/Nov/19

y′re welcom

$${y}'{re}\:{welcom} \\ $$

Commented by mathmax by abdo last updated on 24/Nov/19

f(z)=((πcotan(πz))/(z^2 +z+1))

$${f}\left({z}\right)=\frac{\pi{cotan}\left(\pi{z}\right)}{{z}^{\mathrm{2}} +{z}+\mathrm{1}} \\ $$

Commented by mind is power last updated on 02/Dec/19

yes sir

$$\mathrm{yes}\:\mathrm{sir} \\ $$$$ \\ $$

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