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Question Number 31977 by abdo imad last updated on 17/Mar/18
find the value of Σ_(n=0) ^∞ (1/((2n+1)(2n+3)))
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$
Commented by Tinkutara last updated on 17/Mar/18
(1/2)
$$\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by abdo imad last updated on 18/Mar/18
let put S_n = Σ_(k=0) ^n (1/((2k+1)(2k+3))) we have S_n =(1/2) Σ_(k=0) ^n ( (1/(2k+1)) −(1/(2k+3))) ⇒ 2S_n = Σ_(k=0) ^n (a_k −a_(k+1) ) with a_k =(1/(2k+1)) = a_0 −a_1 +a_1 −a_2 +....+a_n −a_(n+1) =a_0 −a_(n+1) =1−(1/(2n+3)) ⇒ S_n =(1/2) −(1/(2(2n+3))) ⇒lim_(n→∞) S_n =(1/2) .
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}\:{we}\:{have}\: \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(\:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\right)\:\Rightarrow \\ $$$$\mathrm{2}{S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left({a}_{{k}} \:−{a}_{{k}+\mathrm{1}} \right)\:\:{with}\:{a}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\:{a}_{\mathrm{0}} \:−{a}_{\mathrm{1}} \:+{a}_{\mathrm{1}} \:−{a}_{\mathrm{2}} \:+….+{a}_{{n}} \:−{a}_{{n}+\mathrm{1}} ={a}_{\mathrm{0}} \:−{a}_{{n}+\mathrm{1}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}} \\ $$$$\Rightarrow\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{3}\right)}\:\Rightarrow{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$$ \\ $$
Answered by Joel578 last updated on 18/Mar/18
S_k = Σ_(n=0) ^k (1/((2n + 1)(2n + 3))) = (1/2) Σ_(n=0) ^k ((1/(2n + 1)) − (1/(2n + 3))) = (1/2)[(1 − (1/3)) + ((1/3) − (1/5)) + ... + ((1/(2k + 1)) − (1/(2k + 3))) = (1/2)(1 − (1/(2k + 3))) lim_(k→∞) S_k = lim_(k→∞) (1/2)(1 − (1/(2k + 3))) = (1/2)
$${S}_{{k}} \:=\:\underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\left(\mathrm{2}{n}\:+\:\mathrm{3}\right)} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{n}=\mathrm{0}} {\overset{{k}} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{n}\:+\:\mathrm{1}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{n}\:+\:\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\right)\:+\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:−\:\frac{\mathrm{1}}{\mathrm{5}}\right)\:+\:…\:+\:\left(\frac{\mathrm{1}}{\mathrm{2}{k}\:+\:\mathrm{1}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{k}\:+\:\mathrm{3}}\right)\right. \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{2}{k}\:+\:\mathrm{3}}\right) \\ $$$$ \\ $$$$\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:{S}_{{k}} \:=\:\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{2}{k}\:+\:\mathrm{3}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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