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Question Number 155357 by mathlove last updated on 29/Sep/21

Commented by mathlove last updated on 30/Sep/21

$$pleasanswer$$

Answered by qaz last updated on 10/Mar/22

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{(1+\mathrm{n})}^{\frac{1}{{\mathrm{H}}_{\mathrm{n}}}}={\underset{\mathrm{n}\to \mathrm{\infty }}{\lim e}}^{\frac{\ln (1+\mathrm{n})}{\mathrm{lnn}+\gamma }}=\mathrm{e}$$

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