Question Number 181644 by mathlove last updated on 28/Nov/22 | ||
$${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\centerdot\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}}{{xln}\sqrt{{x}}}\right)^{{ln}\sqrt{{x}}} =\mathrm{2}{e}^{\frac{\gamma}{\mathrm{2}}} \: \\ $$$$ \\ $$ | ||
Answered by aleks041103 last updated on 28/Nov/22 | ||
Answered by aleks041103 last updated on 28/Nov/22 | ||
$$\underset{{k}=\mathrm{1}} {\overset{{x}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}=\underset{{k}=\mathrm{1}} {\overset{{x}+\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{k}}\right)\:−\underset{{k}=\mathrm{1}} {\overset{{x}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}= \\ $$$$={H}\left(\mathrm{2}{x}+\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{H}\left({x}+\mathrm{1}\right)= \\ $$$$={H}\left(\mathrm{2}{x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{H}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}= \\ $$$$={H}\left(\mathrm{2}{x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{H}\left({x}\right) \\ $$$${H}\left({x}\right)\rightarrow{ln}\left({x}\right)+\gamma \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{x}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\rightarrow{ln}\left(\mathrm{2}{x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\right)+\frac{\gamma}{\mathrm{2}} \\ $$$$\Rightarrow{L}=\underset{{x}\rightarrow\infty} {{lim}}\left(\frac{{ln}\left(\mathrm{2}\left(\sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left(\sqrt{{x}}\right)+\frac{\gamma}{\mathrm{2}}}{{ln}\left(\sqrt{{x}}\right)}\right)^{{ln}\left(\sqrt{{x}}\right)} = \\ $$$$=\underset{{y}\rightarrow\infty} {{lim}}\left(\frac{{ln}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left({y}\right)+\gamma/\mathrm{2}}{{ln}\left({y}\right)}\right)^{{ln}\left({y}\right)} = \\ $$$$=\underset{{y}\rightarrow\infty} {{lim}}\left(\mathrm{1}+\frac{{ln}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left({y}^{\mathrm{2}} \right)+\gamma/\mathrm{2}}{{ln}\left({y}\right)}\right)^{{ln}\left({y}\right)} = \\ $$$$=\underset{{y}\rightarrow\infty} {{lim}}\left(\mathrm{1}+\frac{{ln}\left(\mathrm{2}+{y}^{−\mathrm{2}} \right)+\gamma/\mathrm{2}}{{ln}\left({y}\right)}\right)^{{ln}\left({y}\right)} = \\ $$$$={exp}\left(\underset{{y}\rightarrow\infty} {{lim}}\left({ln}\left(\mathrm{2}+{y}^{−\mathrm{2}} \right)+\gamma/\mathrm{2}\right)\right)= \\ $$$$={e}^{\frac{\gamma}{\mathrm{2}}+{ln}\left(\mathrm{2}\right)} =\mathrm{2}{e}^{\gamma/\mathrm{2}} \\ $$ | ||
Commented by aleks041103 last updated on 28/Nov/22 | ||
Commented by mathlove last updated on 28/Nov/22 | ||
$${thanks} \\ $$ | ||