Question Number 88360 by jagoll last updated on 10/Apr/20
$$\frac{\mathrm{e}}{\:\sqrt{\mathrm{e}}}\:×\:\frac{\sqrt[{\mathrm{3}\:\:}]{\mathrm{e}}}{\:\sqrt[{\mathrm{4}\:\:}]{\mathrm{e}}}\:×\:\frac{\sqrt[{\mathrm{5}\:\:}]{\mathrm{e}}}{\:\sqrt[{\mathrm{6}\:\:}]{\mathrm{e}}}\:×\:\frac{\sqrt[{\mathrm{7}\:\:}]{\mathrm{e}}}{\:\sqrt[{\mathrm{8}\:\:}]{\mathrm{e}}}×…=? \\ $$
Commented by john santu last updated on 10/Apr/20
![= e^(1−(1/2)+(1/3)−(1/4)+(1/5)−(1/6)+...) [ ln (1+x) = x−(x^2 /2)+(x^3 /3)−(x^4 /4)+... ] −1 < x ≤ 1 , ln (2) = 1−(1/2)+(1/3)−(1/4)+... ] ⇒= e^(ln (2)) = 2.](https://www.tinkutara.com/question/Q88361.png)
$$=\:{e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}+…} \\ $$$$\left[\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}\right)\:=\:\mathrm{x}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}+…\:\right] \\ $$$$\left.−\mathrm{1}\:<\:\mathrm{x}\:\leqslant\:\mathrm{1}\:,\:\mathrm{ln}\:\left(\mathrm{2}\right)\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…\:\right] \\ $$$$\Rightarrow=\:{e}\:^{\mathrm{ln}\:\left(\mathrm{2}\right)} \:=\:\mathrm{2}.\: \\ $$$$\: \\ $$
Commented by jagoll last updated on 10/Apr/20
$$\mathrm{thank}\:\mathrm{you} \\ $$