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Question Number 115405 by mathdave last updated on 25/Sep/20

find the close form of Σ_(n=0) ^∞ (((−1)^n )/((n+1)(n+2)(2n+1)(2n+3)))

$${find}\:{the}\:{close}\:{form}\:{of} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$

Answered by mathmax by abdo last updated on 25/Sep/20

S =lim_(n→+∞) S_n with S_n =Σ_(k=0) ^n (((−1)^k )/((k+1)(k+2)(2k+1)(2k+3))) let decompose F(x) =(1/((x+1)(x+2)(2x+1)(2x+3))) F(x)=(a/(x+1)) +(b/(x+2)) +(c/(2x+1)) +(d/(2x+3)) a =(1/((−1))) =−1 , b =(1/((−1)(−1)(1))) =1 c =(1/(((1/2))((3/2))(2))) =(2/3) , d =(1/((((−1)/2))((1/2))(−2))) =2 ⇒ F(x) =−(1/(x+1)) +(1/(x+2)) +(2/(3(2x+1))) +(2/(2x+3)) ⇒ S_n =−Σ_(k=0) ^n (((−1)^k )/(k+1)) +Σ_(k=0) ^n (((−1)^k )/(k+2)) +(2/3)Σ_(k=0) ^n (((−1)^k )/(2k+1)) +2Σ_(k=0) ^n (((−1)^k )/(2k+3)) we have− Σ_(k=0) ^n (((−1)^k )/(k+1)) =−Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k) =Σ_(k=1) ^(n+1) (((−1)^k )/k)→−ln(2) Σ_(k=0) ^n (((−1)^k )/(k+2)) =Σ_(k=2) ^(n+2) (((−1)^k )/k) =Σ_(k=1) ^(n+2) (((−1)^k )/k) +1→1−ln(2) Σ_(k=0) ^n (((−1)^k )/(2k+1)) →(π/4) Σ_(k=0) ^n (((−1)^k )/(2k+3)) =_(k=p−1) Σ_(p=1) ^(n+1) (((−1)^(p−1) )/(2p+1)) =−Σ_(p=1) ^(n+1) (((−1)^p )/(2p+1)) =−Σ_(p=0) ^(n+1) (((−1)^p )/(2p+1)) +1 =1−(π/4) ⇒ S =lim_(n→+∞) S_n =−ln(2)+1−ln(2)+(2/3).(π/4) +2(1−(π/4)) =1−2ln(2)+(π/6) +2−(π/2) =3−2ln(2)−(π/3)

$$\mathrm{S}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} \:\:\mathrm{with}\:\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right)\left(\mathrm{2k}+\mathrm{1}\right)\left(\mathrm{2k}+\mathrm{3}\right)} \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{2x}+\mathrm{1}\right)\left(\mathrm{2x}+\mathrm{3}\right)} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}+\mathrm{1}}\:+\frac{\mathrm{b}}{\mathrm{x}+\mathrm{2}}\:+\frac{\mathrm{c}}{\mathrm{2x}+\mathrm{1}}\:+\frac{\mathrm{d}}{\mathrm{2x}+\mathrm{3}} \\ $$$$\mathrm{a}\:=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)}\:=−\mathrm{1}\:,\:\mathrm{b}\:=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)\left(\mathrm{1}\right)}\:=\mathrm{1} \\ $$$$\mathrm{c}\:=\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{2}\right)}\:=\frac{\mathrm{2}}{\mathrm{3}}\:,\:\:\mathrm{d}\:=\frac{\mathrm{1}}{\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\mathrm{2}\right)}\:=\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{2x}+\mathrm{1}\right)}\:+\frac{\mathrm{2}}{\mathrm{2x}+\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{S}_{\mathrm{n}} =−\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\mathrm{1}}\:+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{3}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2k}+\mathrm{1}}\:+\mathrm{2}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2k}+\mathrm{3}} \\ $$$$\mathrm{we}\:\mathrm{have}−\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\mathrm{1}}\:=−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}+\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} }{\mathrm{k}}\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}+\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\rightarrow−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\mathrm{2}}\:=\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}+\mathrm{2}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}+\mathrm{2}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\:+\mathrm{1}\rightarrow\mathrm{1}−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2k}+\mathrm{1}}\:\rightarrow\frac{\pi}{\mathrm{4}} \\ $$$$\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2k}+\mathrm{3}}\:=_{\mathrm{k}=\mathrm{p}−\mathrm{1}} \:\:\sum_{\mathrm{p}=\mathrm{1}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{p}−\mathrm{1}} }{\mathrm{2p}+\mathrm{1}}\:=−\sum_{\mathrm{p}=\mathrm{1}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{p}} }{\mathrm{2p}+\mathrm{1}} \\ $$$$=−\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{p}} }{\mathrm{2p}+\mathrm{1}}\:+\mathrm{1}\:=\mathrm{1}−\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{S}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} =−\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{1}−\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{2}}{\mathrm{3}}.\frac{\pi}{\mathrm{4}}\:+\mathrm{2}\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{1}−\mathrm{2ln}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{6}}\:+\mathrm{2}−\frac{\pi}{\mathrm{2}}\:=\mathrm{3}−\mathrm{2ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{3}} \\ $$$$ \\ $$

Commented by mathdave last updated on 25/Sep/20

beautiful approaches but little mistake

$${beautiful}\:{approaches}\:{but}\:{little}\:{mistake} \\ $$

Commented by mathmax by abdo last updated on 25/Sep/20

i think there is no mistakes ...!

$$\mathrm{i}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{mistakes}\:...! \\ $$

Commented by Tawa11 last updated on 06/Sep/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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