Question Number 130733 by bemath last updated on 28/Jan/21
Commented by EDWIN88 last updated on 28/Jan/21
$$\mathrm{10}^{{e}−\mathrm{1}} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Jan/21
$$\mathrm{10}\sqrt{\mathrm{10}\sqrt[{\mathrm{3}}]{\mathrm{10}\sqrt[{\mathrm{4}}]{\mathrm{10}\sqrt[{\mathrm{5}}]{\mathrm{10}..}}}} \\ $$$$=\mathrm{10}^{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}+….} =\mathrm{10}^{{e}−\mathrm{1}} \\ $$
Commented by bemath last updated on 28/Jan/21
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}+\frac{\mathrm{1}}{\mathrm{5}!}+…\:=\:\mathrm{e}−\mathrm{1}\:?\: \\ $$
Commented by Dwaipayan Shikari last updated on 28/Jan/21
$${yes} \\ $$
Commented by bemath last updated on 28/Jan/21
$$\mathrm{e}^{\mathrm{x}} \:=\:\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}+… \\ $$$$\mathrm{e}^{\mathrm{1}} =\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}+\frac{\mathrm{1}}{\mathrm{5}!}+…\: \\ $$$$\mathrm{yes}… \\ $$