Question Number 155357 by mathlove last updated on 29/Sep/21
Commented by mathlove last updated on 30/Sep/21
$${pleas}\:{answer} \\ $$
Answered by qaz last updated on 10/Mar/22
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{n}\right)^{\frac{\mathrm{1}}{\mathrm{H}_{\mathrm{n}} }} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}e}^{\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{n}\right)}{\mathrm{lnn}+\gamma}} =\mathrm{e} \\ $$