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Question Number 144831 by mathdanisur last updated on 29/Jun/21
(3/(1∙2∙3)) + (5/(2∙3∙4)) + (7/(3∙4∙5)) + (9/(4∙5∙6)) + ... ∞=?
$$\frac{\mathrm{3}}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}}\:+\:\frac{\mathrm{5}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}\:+\:\frac{\mathrm{7}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}\:+\:\frac{\mathrm{9}}{\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6}}\:+\:…\:\infty=? \\ $$
Answered by Dwaipayan Shikari last updated on 29/Jun/21
Σ_(n=1) ^∞ ((2n+1)/(n(n+1)(n+2)))=Σ_(n=1) ^∞ (1/(n(n+2)))+(1/((n+1)(n+2))) =(1/2)Σ_(n=1) ^∞ (1/n)−(1/(n+2))+Σ_(n=2) ^∞ (1/n)−(1/(n+1)) =(1/2)(1+(1/2))+(1/2)=(5/4)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Commented by mathdanisur last updated on 29/Jun/21
alot cool thanks Sir
$${alot}\:{cool}\:{thanks}\:{Sir} \\ $$
Answered by mathmax by abdo last updated on 30/Jun/21
S=Σ_(n=1) ^∞ ((2n+1)/(n(n+1)(n+2))) let decompose F(x)=((2x+1)/(x(x+1)(x+2))) ⇒F(x)=(a/x)+(b/(x+1))+(c/(x+2)) a=(1/2) , b=((−1)/((−1)(1)))=1 ,c=((−3)/((−2)(−1)))=−(3/2) ⇒ F(x)=(1/(2x))+(1/(x+1))−(3/(2(x+2))) ⇒ Σ_(k=1) ^n F(k)=(1/2)Σ_(k=1) ^n (1/k)+Σ_(k=1) ^n (1/(k+1))−(3/2)Σ_(k=1) ^n (1/(k+2)) we have Σ_(k=1) ^n (1/k)=H_n Σ_(k=1) ^n (1/(k+1))=Σ_(=2) ^(n+1) (1/k) =H_n −1+(1/(n+1)) Σ_(k=1) ^n (1/(k+2))=Σ_(k=3) ^(n+2) (1/k)=H_n −(3/2)+(1/(n+1))+(1/(n+2)) ⇒ Σ_(k=1) ^n F(k)=(1/2)H_n +H_n −1+(1/(n+1))−(3/2)(H_n −(3/2)+(1/(n+1))+(1/(n+2))) =−1+(1/(n+1))+(9/4)−(3/(2(n+1)))−(3/(2(n+2)))→(9/4)−1=(5/4) ⇒ S=(5/4)
$$\mathrm{S}=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{c}}{\mathrm{x}+\mathrm{2}} \\ $$$$\mathrm{a}=\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mathrm{b}=\frac{−\mathrm{1}}{\left(−\mathrm{1}\right)\left(\mathrm{1}\right)}=\mathrm{1}\:\:,\mathrm{c}=\frac{−\mathrm{3}}{\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)}=−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2x}}+\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{x}+\mathrm{2}\right)}\:\Rightarrow \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{F}\left(\mathrm{k}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}=\mathrm{H}_{\mathrm{n}} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}=\sum_{=\mathrm{2}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{k}}\:=\mathrm{H}_{\mathrm{n}} −\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{2}}=\sum_{\mathrm{k}=\mathrm{3}} ^{\mathrm{n}+\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{k}}=\mathrm{H}_{\mathrm{n}} −\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:\Rightarrow \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{F}\left(\mathrm{k}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{n}} +\mathrm{H}_{\mathrm{n}} −\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{H}_{\mathrm{n}} −\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right) \\ $$$$=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{n}+\mathrm{2}\right)}\rightarrow\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{S}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Commented by mathdanisur last updated on 30/Jun/21
cool thanks Sir
$${cool}\:{thanks}\:{Sir} \\ $$

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