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Question Number 103623 by Lordose last updated on 16/Jul/20
find the sum of the series whose nth term is ((2n−1)/(n(n+1)(n+2)). i have a problem with this and i need help please
$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{series}}\:\boldsymbol{\mathrm{whose}}\:\boldsymbol{\mathrm{nth}} \\ $$$$\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{is}}\:\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}. \\ $$$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{problem}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}} \\ $$$$\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{please}} \\ $$
Answered by OlafThorendsen last updated on 16/Jul/20
find the sum of the series whose nth term is ((2n−1)/(n(n+1)(n+2)). i have a problem with this and i need help please u_n = ((2n−1)/(n(n+1)(n+2))) u_n = −(1/2).(1/n)+(3/(n+1))−(5/2).(1/(n+2)) Σ_(n=1) ^p u_n = −(1/2)Σ_(n=1) ^p (1/n)+3Σ_(n=1) ^p (1/(n+1))−(5/2)Σ_(n=1) ^p (1/(n+2)) Σ_(n=1) ^p u_n = −(1/2)Σ_(n=1) ^p (1/n)+3Σ_(n=2) ^(p+1) (1/n)−(5/2)Σ_(n=3) ^(p+2) (1/n) Σ_(n=1) ^p u_n = −(1/2)(1+(1/2))+3((1/2))+3((1/(p+1)))−(5/2)((1/(p+1))+(1/(p+2))) Σ_(n=1) ^p u_n = (3/4)+(1/2)((1/(p+1)))−(5/2)((1/(p+2))) Σ_(n=1) ^p u_n = (3/4)−((2p+(3/2))/((p+1)(p+2))) and Σ_(n=1) ^∞ u_n = (3/4)
$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{series}}\:\boldsymbol{\mathrm{whose}}\:\boldsymbol{\mathrm{nth}} \\ $$$$\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{is}}\:\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}. \\ $$$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{problem}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}} \\ $$$$\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{please}} \\ $$$${u}_{{n}} \:=\:\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$${u}_{{n}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{{n}}+\frac{\mathrm{3}}{{n}+\mathrm{1}}−\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}{u}_{{n}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}\frac{\mathrm{1}}{{n}}+\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{5}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}{u}_{{n}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}\frac{\mathrm{1}}{{n}}+\mathrm{3}\underset{{n}=\mathrm{2}} {\overset{{p}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{5}}{\mathrm{2}}\underset{{n}=\mathrm{3}} {\overset{{p}+\mathrm{2}} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}{u}_{{n}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3}\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}\right)−\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}+\frac{\mathrm{1}}{{p}+\mathrm{2}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}{u}_{{n}} \:=\:\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}\right)−\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{p}+\mathrm{2}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{p}} {\sum}}{u}_{{n}} \:=\:\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{2}{p}+\frac{\mathrm{3}}{\mathrm{2}}}{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)} \\ $$$$\mathrm{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{u}_{{n}} \:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by Lordose last updated on 16/Jul/20
sir i don′t understand the 4th line can you explain pls
$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{don}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{understand}}\:\boldsymbol{\mathrm{the}}\:\mathrm{4}\boldsymbol{\mathrm{th}}\:\boldsymbol{\mathrm{line}} \\ $$$$\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{explain}}\:\boldsymbol{\mathrm{pls}} \\ $$
Answered by bobhans last updated on 16/Jul/20
Σ_(n=1) ^∞ ((2n−1)/(n(n+1)(n+2))) =Σ_(n=1) ^∞ (3/(n+1))−Σ_(n=1) ^∞ (1/(2n))−Σ_(n=1) ^∞ (5/(2(n+2))) = Σ_(n=2) ^∞ (3/n)−(1/2)Σ_(n=1) ^∞ (1/n)−(5/2)Σ_(n=3) ^∞ (1/n) =Σ_(n=1) ^∞ (3/n)−3−(1/2)Σ_(n=1) ^∞ (1/n)−(5/2){Σ_(n=1) ^∞ (1/n)−1−(1/2)} =−3+((15)/4)+(3−(1/2)−(5/2))Σ_(n=1) ^∞ (1/n) = (3/4) +0 lim_(k→∞) Σ_(n=1) ^k (1/n) = (3/4) + 0 = (3/4)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{3}}{{n}+\mathrm{1}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}{n}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{5}}{\mathrm{2}\left({n}+\mathrm{2}\right)} \\ $$$$=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{{n}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{5}}{\mathrm{2}}\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{{n}}−\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{5}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$=−\mathrm{3}+\frac{\mathrm{15}}{\mathrm{4}}+\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{4}}\:+\mathrm{0}\:\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\:\frac{\mathrm{1}}{{n}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\mathrm{0}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by Lordose last updated on 16/Jul/20
can anyone make me understand pls
$$\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{anyone}}\:\boldsymbol{\mathrm{make}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{understand}}\:\boldsymbol{\mathrm{pls}} \\ $$
Commented by bobhans last updated on 16/Jul/20
what part you don′t understand?
$${what}\:{part}\:{you}\:{don}'{t}\:{understand}? \\ $$

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