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let-S-n-k-1-n-k-2-2k-1-2k-1-1-determine-S-n-interms-of-n-2-find-lim-n-S-n-

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Question Number 48493 by maxmathsup by imad last updated on 24/Nov/18
let S_n =Σ_(k=1) ^n (k^2 /((2k−1)(2k+1))) 1) determine S_n interms of n 2) find lim_(n→+∞) S_n
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{k}^{\mathrm{2}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{S}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 25/Nov/18
1) first let decompose F(x)=(x^2 /((2x−1)(2x+1))) ⇒ F(x)=(x/4)((1/(2x−1)) +(1/(2x+1))) =(1/8)(((2x)/(2x−1)) +((2x)/(2x+1))) =(1/8)(((2x−1+1)/(2x−1)) +((2x+1−1)/(2x+1)))=(1/8)( 1+(1/(2x−1)) +1−(1/(2x+1))) =(1/4) +(1/8)((1/(2x−1)) −(1/(2x+1))) ⇒S_n =(n/4) +(1/8)Σ_(k=1) ^n (1/(2k−1)) −(1/8)Σ_(k=1) ^n (1/(2k+1)) but Σ_(k=1) ^n (1/(2k+1)) =_(k+1=j) Σ_(j=2) ^(n+1) (1/(2j−1)) =Σ_(j=1) ^n (1/(2j−1)) −1 +(1/(2n+1)) ⇒ S_n =(n/4) +(1/8) Σ_(k=1) ^n (1/(2k−1)) −(1/8)( Σ_(k=1) ^n (1/(2k−1)) −1+(1/(2n+1))) =(n/4) +(1/8) −(1/(8(2n+1))) 2) its clear that lim_(n→+∞) S_n =+∞ .
$$\left.\mathrm{1}\right)\:{first}\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{x}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{2}{x}}{\mathrm{2}{x}−\mathrm{1}}\:+\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\:+\frac{\mathrm{2}{x}+\mathrm{1}−\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left(\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right)\:\Rightarrow{S}_{{n}} =\frac{{n}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{8}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$${but}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:=_{{k}+\mathrm{1}={j}} \:\:\:\sum_{{j}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{2}{j}−\mathrm{1}}\:=\sum_{{j}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{j}−\mathrm{1}}\:−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{8}}\left(\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$=\frac{{n}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\left.\mathrm{2}\right)\:{its}\:{clear}\:{that}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =+\infty\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18
1)(k^2 /((2k+1)(2k−1))) =(1/4)[((4k^2 −1+1)/((2k+1)(2k−1)))] =(1/4)[1+(1/2)×(((2k+1)−(2k−1))/((2k+1)(2k−1))] T_k =(1/4)+(1/8)[(1/(2k−1))−(1/(2k+1))] T_1 =(1/4)+(1/8)[(1/1)−(1/3)] T_2 =(1/4)+(1/8)[(1/3)−(1/5)] T_3 =(1/4)+(1/8)[(1/5)−(1/7)] ... ... T_n =(1/4)+(1/8)[(1/(2n−1))−(1/(2n+1))] add them S_n .=(n/4)+(1/8)[1−(1/(2n+1))] when n→∞ S_n =∞+(1/8)[1−0]=∞ plscheck...
$$\left.\mathrm{1}\right)\frac{{k}^{\mathrm{2}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)−\left(\mathrm{2}{k}−\mathrm{1}\right)}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{1}\right.}\right] \\ $$$${T}_{{k}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right] \\ $$$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\right] \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right] \\ $$$${T}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}\right] \\ $$$$… \\ $$$$… \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right] \\ $$$${add}\:{them} \\ $$$${S}_{{n}} .=\frac{{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right] \\ $$$${when}\:{n}\rightarrow\infty\:\: \\ $$$${S}_{{n}} =\infty+\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{1}−\mathrm{0}\right]=\infty \\ $$$${plscheck}… \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 25/Nov/18
correct answer thanks.
$${correct}\:{answer}\:{thanks}. \\ $$
Answered by ajfour last updated on 25/Nov/18
S_n = Σ_(k=1) ^n (k^2 /((2k−1)(2k+1))) = (1/4)Σ ((k[(2k+1)+(2k−1)])/((2k−1)(2k+1))) 4S_n = Σ(k/(2k−1))+Σ(k/(2k+1)) 8S_n = Σ(1+(1/(2k−1)))+Σ(1−(1/(2k+1))) = 2n+((1/1)+(1/3)+..+(1/(2n−1))) −((1/3)+(1/5)+...+(1/(2n+1))) S_n = (n/4)+(1/8)(1−(1/(2n+1))) .
$$\:{S}_{{n}} =\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}\: \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\Sigma\:\frac{{k}\left[\left(\mathrm{2}{k}+\mathrm{1}\right)+\left(\mathrm{2}{k}−\mathrm{1}\right)\right]}{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\:\:\:\mathrm{4}{S}_{{n}} \:=\:\Sigma\frac{{k}}{\mathrm{2}{k}−\mathrm{1}}+\Sigma\frac{{k}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\mathrm{8}{S}_{{n}} \:=\:\Sigma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\right)+\Sigma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2}{n}+\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}}+..+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+…+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:{S}_{{n}} \:=\:\frac{{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 25/Nov/18
correct answer thanks
$${correct}\:{answer}\:{thanks} \\ $$

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