Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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

$$\mathrm{correct}\:\mathrm{answer}\:\mathrm{thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com