Question Number 144483 by alcohol last updated on 25/Jun/21
$$\left({p}_{{n}} \right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right)…\left(\mathrm{1}+\frac{{n}}{{n}^{\mathrm{2}} }\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}{n}\left(\mathrm{2}{n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right) \\ $$$${show}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)−\frac{\mathrm{1}}{\mathrm{12}{n}^{\mathrm{2}} }\left(\mathrm{2}{n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)<{ln}\left({p}_{{n}} \right)<\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right) \\ $$$${hence}\:{find}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({p}_{{n}} \right) \\ $$$$\left.\mathrm{2}\right)\:{show}\:{that}\: \\ $$$${t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\leqslant{ln}\left(\mathrm{1}+{t}\right)\:\leqslant{t},\:\forall{t}>\mathrm{0} \\ $$$${please}\:{help} \\ $$$$ \\ $$
Answered by Olaf_Thorendsen last updated on 26/Jun/21
$$\mathrm{ln}\left(\mathrm{1}+{x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:<\:\mathrm{ln}\left(\mathrm{1}+{x}\right)\:<\:{x}\:\:\:\:\left(\mathrm{1}\right) \\ $$$${p}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$$\mathrm{ln}{p}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{1}\right)\::\:\frac{{k}}{{n}^{\mathrm{2}} }−\frac{{k}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{4}} }\:<\:\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\:<\:\frac{{k}}{{n}^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}}{{n}^{\mathrm{2}} }−\frac{{k}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{4}} }\right)\:<\:\mathrm{ln}{p}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\:<\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{{n}^{\mathrm{2}} } \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} }−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}{n}^{\mathrm{4}} }\:<\:{p}_{{n}} \:<\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)−\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}{n}^{\mathrm{3}} }\:<\:{p}_{{n}} \:<\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right) \\ $$
Commented by alcohol last updated on 26/Jun/21
$${thank}\:{you}\:{brother} \\ $$
Commented by puissant last updated on 07/Jul/21