Question Number 166120 by mathls last updated on 13/Feb/22
Commented by mathls last updated on 13/Feb/22
$${plese}\:{help}\:{me} \\ $$
Commented by mathls last updated on 13/Feb/22
$$? \\ $$
Answered by amin96 last updated on 13/Feb/22
![S=Σ_(n=1) ^(2000) ((1/(n+1))/(Π_(m=1) ^n (1+(1/(m+1)))))=Σ_(n=1) ^(2000) ((1/(n+1))/(Π_(m=1) ^n ((m+2)/(m+1)))) Π_(m=1) ^n ((m+2)/(m+1))=(3/2)×(4/3)×(5/4).....((n+2)/(n+1))=((n+2)/2) S=Σ_(n=1) ^(2000) ((1/(n+1))/((n+2)/2))=^ Σ_(n=1) ^(2000) (2/((n+1)(n+2)))= =2Σ_(n=1) ^(2000) ((n+2−(n+1))/((n+1)(n+2)))=2[Σ_(n=1) ^(2000) (1/(n+1))−Σ_(n=1) ^(2000) (1/(n+2))]= =2[(1/2)+(1/3)+…+(1/(2001))−((1/3)+(1/4)+…+(1/(2001)))]= =1](Q166133.png)
$$\boldsymbol{\mathrm{S}}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\frac{\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{1}}}{\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{m}}+\mathrm{1}}\right)}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\frac{\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{1}}}{\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\frac{\boldsymbol{\mathrm{m}}+\mathrm{2}}{\boldsymbol{\mathrm{m}}+\mathrm{1}}} \\ $$$$\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\frac{\boldsymbol{\mathrm{m}}+\mathrm{2}}{\boldsymbol{\mathrm{m}}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{5}}{\mathrm{4}}.....\frac{\boldsymbol{\mathrm{n}}+\mathrm{2}}{\boldsymbol{\mathrm{n}}+\mathrm{1}}=\frac{\boldsymbol{\mathrm{n}}+\mathrm{2}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{S}}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\frac{\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{1}}}{\frac{\boldsymbol{\mathrm{n}}+\mathrm{2}}{\mathrm{2}}}\hat {=}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\frac{\mathrm{2}}{\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)}= \\ $$$$=\mathrm{2}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\frac{\boldsymbol{\mathrm{n}}+\mathrm{2}−\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)}{\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)}=\mathrm{2}\left[\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{1}}−\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\mathrm{2000}} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{2}}\right]= \\ $$$$=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\ldots+\frac{\mathrm{1}}{\mathrm{2001}}−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\ldots+\frac{\mathrm{1}}{\mathrm{2001}}\right)\right]= \\ $$$$=\mathrm{1} \\ $$