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Question Number 113789 by Khalmohmmad last updated on 15/Sep/20
Σ_(n=1) ^∝ (3/(n(n+3)))=?
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\propto} {\sum}}\frac{\mathrm{3}}{\mathrm{n}\left(\mathrm{n}+\mathrm{3}\right)}=? \\ $$
Answered by aleks041103 last updated on 15/Sep/20
(3/(n(n+3)))=(((n+3)−n)/(n(n+3)))=(1/n)−(1/(n+3)) Σ_(n=1) ^∞ (3/(n(n+3)))=Σ_(n=1) ^∞ ((1/n)−(1/(n+3)))= =(Σ_(n=1) ^∞ (1/n))−(Σ_(n=1) ^∞ (1/(n+3)))= =(Σ_(n=1) ^∞ (1/n))−(Σ_(n=4) ^∞ (1/n))= =(Σ_(n=1) ^3 (1/n) + Σ_(n=4) ^∞ (1/n))−(Σ_(n=4) ^∞ (1/n))= =Σ_(n=1) ^3 (1/n)=(1/1)+(1/2)+(1/3)=((11)/6) ⇒Σ_(n=1) ^∞ (3/(n(n+3)))=((11)/6)
$$\frac{\mathrm{3}}{{n}\left({n}+\mathrm{3}\right)}=\frac{\left({n}+\mathrm{3}\right)−{n}}{{n}\left({n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{{n}\left({n}+\mathrm{3}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)= \\ $$$$=\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\right)−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)= \\ $$$$=\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\right)−\left(\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\right)= \\ $$$$=\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{{n}}\:+\:\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\right)−\left(\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\right)= \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{{n}}=\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{{n}\left({n}+\mathrm{3}\right)}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$
Answered by Dwaipayan Shikari last updated on 15/Sep/20
Σ_(n=1) ^∞ (((n+2))/(n(n+2)(n+3)))=Σ^∞ (1/(n+2))−(1/(n+3))+Σ^∞ (2/(n(n+2)(n+3))) =((1/3)−0)+2Σ^∞ ((n+1)/(n(n+1)(n+2)(n+3))) =(1/3)+2Σ^∞ (1/((n+1)(n+2)(n+3)))+2Σ^∞ (1/(n(n+1)(n+2)(n+3))) =(1/3)+Σ^∞ ((n+3−n−1)/((n+1)(n+2)(n+3)))+(2/3)Σ^∞ ((n+3−n)/(n(n+1)(n+2)(n+3))) =(1/3)+Σ^∞ (1/((n+1)(n+2)))−Σ^∞ (1/((n+2)(n+3)))+(2/3)Σ^∞ (1/(n(n+1)(n+2)))−(2/3)Σ^∞ (1/((n+1)(n+2)(n+3))) =(1/3)+(1/2)−(1/3)+(1/3)Σ^∞ (1/(n(n+1)))−(1/((n+1)(n+2)))−(1/3)Σ^∞ (1/((n+1)(n+2)))−(1/((n+2)(n+3))) =(1/2)+(1/3)(1−(1/2))−(1/3)((1/2))+(1/3).(1/3) =(1/2)+(1/9)=((11)/(18)) So Σ_(n=1) ^∞ (1/(n(n+3)))=((11)/(18)) so,Σ_(n=1) ^∞ (3/(n(n+3)))=((11)/6)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{2}\right)}{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}=\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}+\overset{\infty} {\sum}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{0}\right)+\mathrm{2}\overset{\infty} {\sum}\frac{{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}+\mathrm{2}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\overset{\infty} {\sum}\frac{{n}+\mathrm{3}−{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}+\frac{\mathrm{2}}{\mathrm{3}}\overset{\infty} {\sum}\frac{{n}+\mathrm{3}−{n}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}+\frac{\mathrm{2}}{\mathrm{3}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\frac{\mathrm{2}}{\mathrm{3}}\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{3}}\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{11}}{\mathrm{18}} \\ $$$${So} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{3}\right)}=\frac{\mathrm{11}}{\mathrm{18}}\:\:\:\:{so},\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{{n}\left({n}+\mathrm{3}\right)}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$ \\ $$

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