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Question Number 60676 by maxmathsup by imad last updated on 24/May/19

let S_n =Σ_(k=1) ^n sin^2 (((kπ)/n^2 )) find lim_(n→∞) S_n

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}^{\mathrm{2}} }\right)\:\:\:\:\:{find}\:{lim}_{{n}\rightarrow\infty} \:\:{S}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 31/May/19

we have S_n =Σ_(k=0) ^n sin^2 (((kπ)/n^2 )) =Σ_(k=0) ^n ((1−cos(((2kπ)/n^2 )))/2) =(1/2)Σ_(k=0) ^n (1)−(1/2)Σ_(k=0) ^n cos(((2kπ)/n^2 )) =((n+1)/2) −(1/2)Σ_(k=0) ^n cos(((2kπ)/n^2 )) Σ_(k=0) ^n cos(((2kπ)/n^2 )) =Re(Σ_(k=0) ^n e^(2i((kπ)/n^2 )) ) but Σ_(k=0) ^n e^(2i((kπ)/n^2 )) =Σ_(k=0) ^n (e^((2iπ)/n^2 ) )^k =((1−(e^((2iπ)/n^2 ) )^(n+1) )/(1−e^((2iπ)/n^2 ) )) =((1−e^((2iπ(n+1))/n^2 ) )/(1−e^((2iπ)/n^2 ) )) =((1−cos(((2π(n+1))/n^2 ))−isin(((2π(n+1))/n^2 )))/(1−cos(((2π)/n^2 ))−isin(((2π)/n^2 )))) =((2sin^2 (((π(n+1))/n^2 ))−2i sin(((π(n+1))/(n^2 )))cos(((π(n+1))/n^2 )))/(2sin^2 ((π/n^2 ))−2i sin((π/n^2 ))cos((π/n^2 )))) =((−i sin(((π(n+1))/n^2 )){cos(((π(n+1))/n^2 ))+isin(((π(n+1))/n^2 )))/(−isin((π/n^2 )){cos((π/n^2 ))+isin((π/n^2 )))) =((sin(((π(n+1))/n^2 )))/(sin((π/n^2 )))) (e^((iπ(n+1))/n^2 ) /e^((iπ)/n^2 ) ) =((sin(((π(n+1))/n^2 )))/(sin((π/n^2 )))) e^((iπ)/n) ⇒ Σ_(k=0) ^n cos(((2kπ)/n^2 )) = cos((π/n))((sin((((n+1)π)/n^2 )))/(sin((π/n^2 )))) ⇒ S_n =((n+1)/2) −(1/2) cos((π/n))((sin((((n+1)π)/n^2 )))/(sin((π/n^2 )))) .

$${we}\:{have}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}^{\mathrm{2}} }\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}^{\mathrm{2}} }\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}} {cos}\left(\frac{\mathrm{2}{k}\pi}{{n}^{\mathrm{2}} }\right)\:=\frac{{n}+\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}^{\mathrm{2}} }\right) \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}^{\mathrm{2}} }\right)\:={Re}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{e}^{\mathrm{2}{i}\frac{{k}\pi}{{n}^{\mathrm{2}} }} \right)\:\:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{e}^{\mathrm{2}{i}\frac{{k}\pi}{{n}^{\mathrm{2}} }} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\left({e}^{\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} \right)^{{k}} \:=\frac{\mathrm{1}−\left({e}^{\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} }\:\:=\frac{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }} }{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} } \\ $$$$=\frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)−{isin}\left(\frac{\mathrm{2}\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)}{\mathrm{1}−{cos}\left(\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\right)−{isin}\left(\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)−\mathrm{2}{i}\:{sin}\left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} \:}\right){cos}\left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{{n}^{\mathrm{2}} }\right)−\mathrm{2}{i}\:{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right){cos}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)} \\ $$$$=\frac{−{i}\:{sin}\left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)\left\{{cos}\left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)+{isin}\left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)\right.}{−{isin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\left\{{cos}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)+{isin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\right.} \\ $$$$=\frac{{sin}\left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\:\frac{{e}^{\frac{{i}\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }} }{{e}^{\frac{{i}\pi}{{n}^{\mathrm{2}} }} }\:=\frac{{sin}\left(\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right)}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\:{e}^{\frac{{i}\pi}{{n}}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}^{\mathrm{2}} }\right)\:=\:{cos}\left(\frac{\pi}{{n}}\right)\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}^{\mathrm{2}} }\right)}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\:\Rightarrow \\ $$$${S}_{{n}} =\frac{{n}+\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{cos}\left(\frac{\pi}{{n}}\right)\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}^{\mathrm{2}} }\right)}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\:. \\ $$

Commented by maxmathsup by imad last updated on 31/May/19

we have cos((π/n))∼1−(π^2 /(2n^2 )) , sin((((n+1)π)/n^2 ))∼(((n+1)π)/n^2 ) , sin((π/n^2 ))∼(π/n^2 ) (n∈v(∞)) S_n ∼ ((n+1)/2) −(1/2)(1−(π^2 /n^2 )) (n+1) =((n+1)/2) −((n+1)/2) +(((n+1)π^2 )/(2n^2 )) ⇒ S_n ∼(((n+1)π^2 )/n^2 ) ⇒ lim_(n→∞) S_n =0 .

$${we}\:{have}\:\:{cos}\left(\frac{\pi}{{n}}\right)\sim\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} }\:\:\:,\:\:\:\:{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}^{\mathrm{2}} }\right)\sim\frac{\left({n}+\mathrm{1}\right)\pi}{{n}^{\mathrm{2}} }\:\:,\:{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\sim\frac{\pi}{{n}^{\mathrm{2}} }\:\:\left({n}\in{v}\left(\infty\right)\right) \\ $$$${S}_{{n}} \sim\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\:\left({n}+\mathrm{1}\right)\:=\frac{{n}+\mathrm{1}}{\mathrm{2}}\:−\frac{{n}+\mathrm{1}}{\mathrm{2}}\:+\frac{\left({n}+\mathrm{1}\right)\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} }\:\Rightarrow\:{S}_{{n}} \sim\frac{\left({n}+\mathrm{1}\right)\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \:\:{S}_{{n}} =\mathrm{0}\:. \\ $$

Answered by Smail last updated on 24/May/19

S_n =Σ_(k=1) ^n sin^2 (((kπ)/n^2 ))=Σ_(k=1) ^n (((1−cos(2((kπ)/n^2 )))/2)) =(1/2)Σ_(k=1) ^n −(1/2)Σ_(k=1) ^n cos(((2kπ)/n^2 )) =(n/2)−(1/2)Re(Σ_(k=1) ^n e^((2ikπ)/n^2 ) ) Σ_(k=1) ^n e^((2ikπ)/n^2 ) =Σ_(k=1) ^n (e^((2iπ)/n^2 ) )^k =((e^((2iπ)/n^2 ) (1−e^((2inπ)/n^2 ) ))/(1−e^((2iπ)/n^2 ) )) =((1−e^((2iπ)/n) )/(e^(−((2iπ)/n^2 )) −1))=((1−e^((2iπ)/n) )/(cos(((2π)/n^2 ))−1−isin(((2π)/n^2 )))) =((1−e^((2iπ)/n) )/(−2sin^2 ((π/n^2 ))−2isin((π/n^2 ))cos((π/n^2 )))) =((e^((2iπ)/n) −1)/(2isin((π/n^2 ))(cos((π/n^2 ))−isin((π/n^2 ))))) =((e^((2iπ)/n) −1)/(2isin((π/n^2 ))e^(−i(π/n^2 )) ))=((ie^((iπ)/n^2 ) (1−e^(2iπ/n) ))/(2sin(π/n^2 ))) =((ie^(i(π/n^2 )) (1−cos(2π/n)−isin(2π/n)))/(2sin(π/n^2 ))) =((ie^(iπ/n^2 ) (2sin^2 (π/n)−2isin(π/n)cos(π/n))/(2sin(π/n^2 ))) =ie^(iπ/n^2 ) ((2isin(π/n)(−isin(π/n)−cos(π/n)))/(2sin(π/n^2 ))) =((sin(π/n))/(sin(π/n^2 )))(e^(iπ/n^2 ) ×e^(iπ/n) )=((sin(π/n))/(sin(π/n^2 )))e^(i((π(n+1))/n^2 )) Re(((sin(π/n))/(sin(π/n^2 )))e^(i((π(n+1))/n^2 )) )=((sin(π/n))/(sin(π/n^2 )))cos(((n+1)/n^2 )π) Σ_(k=1) ^n sin^2 (((kπ)/n^2 ))=(n/2)−((sin(π/n))/(2sin(π/n^2 )))cos(((n+1)/n^2 )π)

$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}^{\mathrm{2}} }\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}−{cos}\left(\mathrm{2}\frac{{k}\pi}{{n}^{\mathrm{2}} }\right)}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}^{\mathrm{2}} }\right) \\ $$$$=\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{Re}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{\frac{\mathrm{2}{ik}\pi}{{n}^{\mathrm{2}} }} \right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{\frac{\mathrm{2}{ik}\pi}{{n}^{\mathrm{2}} }} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({e}^{\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} \right)^{{k}} =\frac{{e}^{\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} \left(\mathrm{1}−{e}^{\frac{\mathrm{2}{in}\pi}{{n}^{\mathrm{2}} }} \right)}{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} } \\ $$$$=\frac{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} }{{e}^{−\frac{\mathrm{2}{i}\pi}{{n}^{\mathrm{2}} }} −\mathrm{1}}=\frac{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} }{{cos}\left(\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\right)−\mathrm{1}−{isin}\left(\frac{\mathrm{2}\pi}{{n}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} }{−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{{n}^{\mathrm{2}} }\right)−\mathrm{2}{isin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right){cos}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)} \\ $$$$=\frac{{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} −\mathrm{1}}{\mathrm{2}{isin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\left({cos}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)−{isin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\right)} \\ $$$$=\frac{{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} −\mathrm{1}}{\mathrm{2}{isin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right){e}^{−{i}\frac{\pi}{{n}^{\mathrm{2}} }} }=\frac{{ie}^{\frac{{i}\pi}{{n}^{\mathrm{2}} }} \left(\mathrm{1}−{e}^{\mathrm{2}{i}\pi/{n}} \right)}{\mathrm{2}{sin}\left(\pi/{n}^{\mathrm{2}} \right)} \\ $$$$=\frac{{ie}^{{i}\frac{\pi}{{n}^{\mathrm{2}} }} \left(\mathrm{1}−{cos}\left(\mathrm{2}\pi/{n}\right)−{isin}\left(\mathrm{2}\pi/{n}\right)\right)}{\mathrm{2}{sin}\left(\pi/{n}^{\mathrm{2}} \right)} \\ $$$$=\frac{{ie}^{{i}\pi/{n}^{\mathrm{2}} } \left(\mathrm{2}{sin}^{\mathrm{2}} \left(\pi/{n}\right)−\mathrm{2}{isin}\left(\pi/{n}\right){cos}\left(\pi/{n}\right)\right.}{\mathrm{2}{sin}\left(\pi/{n}^{\mathrm{2}} \right)} \\ $$$$={ie}^{{i}\pi/{n}^{\mathrm{2}} } \frac{\mathrm{2}{isin}\left(\pi/{n}\right)\left(−{isin}\left(\pi/{n}\right)−{cos}\left(\pi/{n}\right)\right)}{\mathrm{2}{sin}\left(\pi/{n}^{\mathrm{2}} \right)} \\ $$$$=\frac{{sin}\left(\pi/{n}\right)}{{sin}\left(\pi/{n}^{\mathrm{2}} \right)}\left({e}^{{i}\pi/{n}^{\mathrm{2}} } ×{e}^{{i}\pi/{n}} \right)=\frac{{sin}\left(\pi/{n}\right)}{{sin}\left(\pi/{n}^{\mathrm{2}} \right)}{e}^{{i}\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }} \\ $$$${Re}\left(\frac{{sin}\left(\pi/{n}\right)}{{sin}\left(\pi/{n}^{\mathrm{2}} \right)}{e}^{{i}\frac{\pi\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }} \right)=\frac{{sin}\left(\pi/{n}\right)}{{sin}\left(\pi/{n}^{\mathrm{2}} \right)}{cos}\left(\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }\pi\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}^{\mathrm{2}} }\right)=\frac{{n}}{\mathrm{2}}−\frac{{sin}\left(\pi/{n}\right)}{\mathrm{2}{sin}\left(\pi/{n}^{\mathrm{2}} \right)}{cos}\left(\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }\pi\right) \\ $$

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