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Question Number 103633 by Lordose last updated on 16/Jul/20
The question is Σ_(n=1) ^∞ (((2n−1)/(n(n+1)(n+2)))=...
$$ \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{is}} \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}\right)=… \\ $$
Commented by bobhans last updated on 16/Jul/20
well well
$${well}\:{well}\: \\ $$
Answered by Dwaipayan Shikari last updated on 16/Jul/20
S=Σ^∞ (((n+n+2−3)/(n(n+1)(n+2))))=Σ^∞ (1/((n+1)(n+2)))+Σ^∞ (1/(n(n+1)))−3S_1 S=Σ^∞ (1/(n+1))−(1/(n+2))+Σ^∞ (1/n)−(1/(n+1))−3S_1 =(3/2)−3S_1 S_1 =(1/2)Σ^∞ ((2+n−n)/(n(n+1)(n+2)))=(1/2)Σ^∞ (1/n)−(1/(n+1))−(1/2)Σ^∞ (1/(n+1))−(1/(n+2)) S_1 =(1/2)−(1/4)=(1/4) { Σ^∞ (1/n)−(1/(n+1))=lim_(n→∞) 1−(1/(n+1))=1 S=(3/2)−(3/4)=(3/4) {Σ^∞ (1/(n+1))−(1/(n+2))=lim_(n→∞) (1/2)−(1/(n+2))=(1/2)
$${S}=\overset{\infty} {\sum}\left(\frac{{n}+{n}+\mathrm{2}−\mathrm{3}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right)=\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}+\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\mathrm{3}{S}_{\mathrm{1}} \\ $$$${S}=\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}+\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}−\mathrm{3}{S}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3}{S}_{\mathrm{1}} \\ $$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\sum}\frac{\mathrm{2}+{n}−{n}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\left\{\:\:\:\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{1}\right. \\ $$$${S}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\left\{\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\right. \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 16/Jul/20
not correct
$$\mathrm{not}\:\mathrm{correct} \\ $$
Commented by Dwaipayan Shikari last updated on 16/Jul/20
kindly can you show my mistake?
$$\mathrm{kindly}\:\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:\mathrm{my}\:\mathrm{mistake}? \\ $$
Answered by Worm_Tail last updated on 16/Jul/20
Σ_(n=1) ^∞ (((2n−1)/(n(n+1)(n+2)))=A pfd A=Σ((3/(n+1))−(5/(2(n+2)))−(1/(2n))) A= 3Σ((1/(n+1)))−(5/2)Σ((1/(n+2)))−(1/2)Σ((1/n)) A= 3((1/2)+Σ_(n=2) ((1/(n+1))))−(5/2)Σ_(n=1) ((1/(n+2)))−(1/2)(1+(1/2)+Σ_(n=3) ((1/n))) Σ_(n=2) ((1/(n+1)))=Σ_(n=1) ((1/(n+2)))=Σ_(n=3) ((1/n))=s A= 3((1/2)+s)−(5/2)s−(1/2)(1+(1/2)+s) A= (3/2)+3s−(5/2)s−(1/2)−(1/4)−(1/2)s A= 3s−(5/2)s−(1/2)s+(3/4) A= (3/4)
$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}\right)={A} \\ $$$$\:\:\:\:{pfd} \\ $$$$\:\:\:{A}=\Sigma\left(\frac{\mathrm{3}}{{n}+\mathrm{1}}−\frac{\mathrm{5}}{\mathrm{2}\left({n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\:\:\:\:{A}=\:\:\:\mathrm{3}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\frac{\mathrm{5}}{\mathrm{2}}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\Sigma\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:\:\:{A}=\:\:\:\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}+\underset{{n}=\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\right)−\frac{\mathrm{5}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\underset{{n}=\mathrm{3}} {\sum}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$\:\:\:\:\underset{{n}=\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)=\underset{{n}=\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)=\underset{{n}=\mathrm{3}} {\sum}\left(\frac{\mathrm{1}}{{n}}\right)={s} \\ $$$$\:\:\:\:{A}=\:\:\:\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}+{s}\right)−\frac{\mathrm{5}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+{s}\right) \\ $$$$\:\:\:\:{A}=\:\:\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{3}{s}−\frac{\mathrm{5}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{s} \\ $$$$\:\:\:\:{A}=\:\:\:\mathrm{3}{s}−\frac{\mathrm{5}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}{s}+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\:{A}=\:\:\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\: \\ $$
Commented by mathmax by abdo last updated on 16/Jul/20
correct answer thanks
$$\mathrm{correct}\:\mathrm{answer}\:\mathrm{thanks} \\ $$

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