Arithmetic Questions

Question Number 103633 by Lordose last updated on 16/Jul/20

$$\\$$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{is}} \\$$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}\right)=... \\$$

Commented by bobhans last updated on 16/Jul/20

$${well}\:{well}\: \\$$

Answered by Dwaipayan Shikari last updated on 16/Jul/20

$${S}=\overset{\infty} {\sum}\left(\frac{{n}+{n}+\mathrm{2}−\mathrm{3}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right)=\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}+\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\mathrm{3}{S}_{\mathrm{1}} \\$$$${S}=\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}+\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}−\mathrm{3}{S}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3}{S}_{\mathrm{1}} \\$$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\sum}\frac{\mathrm{2}+{n}−{n}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}} \\$$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\left\{\:\:\:\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{1}\right. \\$$$${S}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\left\{\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\right. \\$$$$\\$$$$\\$$

Commented by mathmax by abdo last updated on 16/Jul/20

$$\mathrm{not}\:\mathrm{correct} \\$$

Commented by Dwaipayan Shikari last updated on 16/Jul/20

$$\mathrm{kindly}\:\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:\mathrm{my}\:\mathrm{mistake}? \\$$

Answered by Worm_Tail last updated on 16/Jul/20

$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right.}\right)={A} \\$$$$\:\:\:\:{pfd} \\$$$$\:\:\:{A}=\Sigma\left(\frac{\mathrm{3}}{{n}+\mathrm{1}}−\frac{\mathrm{5}}{\mathrm{2}\left({n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\$$$$\:\:\:\:{A}=\:\:\:\mathrm{3}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\frac{\mathrm{5}}{\mathrm{2}}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\Sigma\left(\frac{\mathrm{1}}{{n}}\right) \\$$$$\:\:\:\:{A}=\:\:\:\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}+\underset{{n}=\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\right)−\frac{\mathrm{5}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\underset{{n}=\mathrm{3}} {\sum}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\$$$$\:\:\:\:\underset{{n}=\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)=\underset{{n}=\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)=\underset{{n}=\mathrm{3}} {\sum}\left(\frac{\mathrm{1}}{{n}}\right)={s} \\$$$$\:\:\:\:{A}=\:\:\:\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}+{s}\right)−\frac{\mathrm{5}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+{s}\right) \\$$$$\:\:\:\:{A}=\:\:\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{3}{s}−\frac{\mathrm{5}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{s} \\$$$$\:\:\:\:{A}=\:\:\:\mathrm{3}{s}−\frac{\mathrm{5}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}{s}+\frac{\mathrm{3}}{\mathrm{4}} \\$$$$\:\:\:\:{A}=\:\:\:\frac{\mathrm{3}}{\mathrm{4}} \\$$$$\:\:\:\:\:\: \\$$

Commented by mathmax by abdo last updated on 16/Jul/20

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