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Question Number 80036 by jagoll last updated on 30/Jan/20

Σ_(n=1) ^∞ (1/((n+1)(n+2)(n+3)))=

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}=\: \\ $$

Commented by john santu last updated on 30/Jan/20

lim_(p→∞) Σ_(n=1) ^p (((n+3)−(n+1))/(2(n+1)(n+2)(n+3))) = (1/2)lim_(p→∞) Σ_(n=1 ) ^p (1/((n+1)(n+2)))−(1/((n+2)(n+3))) =(1/2){(1/(2.3))−(1/(3.4))+(1/(3.4))−(1/(4.5))+(1/(4.5))−(1/(5.6))+...} the last term equat to zero , all in between terms are cancelled ∴ Σ_(n=1) ^∞ (1/((n+1)(n+2)(n+3))) = (1/(12))

$$\underset{\mathrm{p}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{p}} {\sum}}\:\frac{\left(\mathrm{n}+\mathrm{3}\right)−\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{p}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{n}=\mathrm{1}\:} {\overset{\mathrm{p}} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{5}.\mathrm{6}}+...\right\} \\ $$$$\mathrm{the}\:\mathrm{last}\:\mathrm{term}\:\mathrm{equat}\:\mathrm{to}\:\mathrm{zero}\:,\: \\ $$$$\mathrm{all}\:\mathrm{in}\:\mathrm{between}\:\mathrm{terms}\:\mathrm{are}\:\mathrm{cancelled} \\ $$$$\therefore\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented by mr W last updated on 30/Jan/20

the question was originally Σ_(n=1) ^∞ (n/((n+1)(n+2)(n+3)))= which has the result (1/4).

$${the}\:{question}\:{was}\:{originally} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}=\: \\ $$$${which}\:{has}\:{the}\:{result}\:\frac{\mathrm{1}}{\mathrm{4}}. \\ $$

Commented by jagoll last updated on 31/Jan/20

waw...i have two solution (1) Σ_(n=1) ^∞ (n/((n+1)(n+2)(n+3)))=(1/4) (2) Σ_(n=1) ^∞ (1/((n+1)(n+2)(n+3))) = (1/(12)) thanks you

$${waw}...{i}\:{have}\:{two}\:{solution} \\ $$$$\left(\mathrm{1}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${thanks}\:{you} \\ $$

Answered by Kamel Kamel last updated on 30/Jan/20

(n/((n+1)(n+2)(n+3)))=(A/(n+1))+(B/(n+2))+(C/(n+3))...(E) (E)×(n+1),n=−1⇒A=−(1/2) (E)×(n+2),n=−2⇒B=2 (E)×(n+3),n=−3⇒C=−(3/2) ∴ (n/((n+1)(n+2)(n+3)))=(1/2)((1/(n+2))−(1/(n+1)))−(3/2)((1/(n+3))−(1/(n+2))) ∴ Σ_(n=1) ^(+∞) (n/((n+1)(n+2)(n+3)))=(1/2)(Σ_(n=1) ^(+∞) ((1/(n+2))−(1/(n+1)))−3Σ_(n=1) ^(+∞) ((1/(n+3))−(1/(n+2)))) =(1/2)(((1/3)−(1/2)+(1/4)−(1/3)+...)−3((1/4)−(1/3)+(1/5)−(1/4)+...)) =(1/2)(−(1/2)+1)=(1/4)

$$\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)}=\frac{\mathrm{A}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{B}}{\mathrm{n}+\mathrm{2}}+\frac{\mathrm{C}}{\mathrm{n}+\mathrm{3}}...\left(\mathrm{E}\right) \\ $$$$\left(\mathrm{E}\right)×\left(\mathrm{n}+\mathrm{1}\right),\mathrm{n}=−\mathrm{1}\Rightarrow\mathrm{A}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{E}\right)×\left(\mathrm{n}+\mathrm{2}\right),\mathrm{n}=−\mathrm{2}\Rightarrow\mathrm{B}=\mathrm{2} \\ $$$$\left(\mathrm{E}\right)×\left(\mathrm{n}+\mathrm{3}\right),\mathrm{n}=−\mathrm{3}\Rightarrow\mathrm{C}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\therefore\:\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right) \\ $$$$\therefore\:\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)−\mathrm{3}\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}+...\right)−\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{4}}+...\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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