Question Number 220069 by hardmath last updated on 04/May/25
$$\mathrm{Let}\:\mathrm{be}\:\:\:\left(\mathrm{H}_{\boldsymbol{\mathrm{n}}} \right)_{\boldsymbol{\mathrm{n}}\geqslant\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{H}_{\boldsymbol{\mathrm{n}}} \:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{2}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} } \:\left(\sqrt[{\boldsymbol{\mathrm{n}}+\mathrm{1}}]{\left(\mathrm{n}+\mathrm{1}\right)!}\:−\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\:\right)\:\mathrm{sin}\:\frac{\pi}{\mathrm{n}^{\mathrm{2}} }\:=\:? \\ $$
Answered by MrGaster last updated on 05/May/25
$${H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\approx\mathrm{ln}\:{n}+\gamma \\ $$$${e}^{\mathrm{2}{H}_{{n}} } \approx{e}^{\mathrm{2}} \gamma{n}^{\mathrm{2}} \\ $$$$\sqrt[{{n}}]{{n}!}\approx\frac{{n}}{{e}}\left(\mathrm{1}+\frac{\mathrm{ln}\left(\mathrm{2}\pi{n}\right)}{\mathrm{2}{n}}\right) \\ $$$$\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!}−\sqrt[{{n}}]{{n}!}\approx\frac{\mathrm{1}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}{en}} \\ $$$$\mathrm{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\approx\frac{\pi}{{n}^{\mathrm{2}} } \\ $$$${e}^{\mathrm{2}{H}_{{n}} } \left(\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!}−\sqrt[{{n}}]{{n}!}\right)\mathrm{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\approx{e}^{\mathrm{2}} \gamma{n}^{\mathrm{2}} \centerdot\frac{\mathrm{1}}{{e}}\centerdot\frac{\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }={e}^{\mathrm{2}\gamma−\mathrm{1}} \pi \\ $$
Commented by hardmath last updated on 05/May/25
$$ \\ $$Amazing solution as always, thank you very much my valuable professor