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1-1-2-1-3-4-1-5-6-1-39-40-




Question Number 160948 by Rustambek last updated on 09/Dec/21
(1/(1∙2))+(1/(3∙4))+(1/(5∙6))+...+(1/(39∙40))=?
$$\frac{\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}\centerdot\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{6}}+…+\frac{\mathrm{1}}{\mathrm{39}\centerdot\mathrm{40}}=? \\ $$
Answered by puissant last updated on 09/Dec/21
Ω=Σ_(k=1) ^(39) (1/(k(k+1))) = Σ_(k=1) ^(39) {(1/k)−(1/(k+1))}  = 1−(1/2)+(1/2)−(1/3)+.......−(1/(39))+(1/(39))−(1/(40))  = 1−(1/(40))     ⇒    Ω = ((39)/(40))....
$$\Omega=\underset{{k}=\mathrm{1}} {\overset{\mathrm{39}} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{39}} {\sum}}\left\{\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right\} \\ $$$$=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+…….−\frac{\mathrm{1}}{\mathrm{39}}+\frac{\mathrm{1}}{\mathrm{39}}−\frac{\mathrm{1}}{\mathrm{40}} \\ $$$$=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{40}}\:\:\:\:\:\Rightarrow\:\:\:\:\Omega\:=\:\frac{\mathrm{39}}{\mathrm{40}}…. \\ $$

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