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