Question Number 174389 by Best1 last updated on 31/Jul/22
$${what}\:{is}\:\:{least}\:{upper}\:{bound}\:{of}\:{the} \\ $$$${sequence}\:\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right\}_{{n}=\mathrm{1}} ^{\infty} ? \\ $$$${A}.\mathrm{0}\:\:\:\:{B}.{ln}\mathrm{2}\:\:\:\:\:\:{C}.{ln}\mathrm{5}\:\:\:\:\:\:\:{D}.\mathrm{2}{ln}\mathrm{2}\: \\ $$
Answered by Gazella thomsonii last updated on 31/Jul/22
![Let′s think of a Sequence A_k Function as f(z) and Find Root f^((1)) (z)=0 for find inf [f(z),z_0 ] so , f(z)=(1+(1/z))ln(1+(1/z)) ((df(z))/dz)=−((1/z))^2 ln(1+(1/z))−((1+(1/z))/(1+(1/z)))∙((d )/dz)(1+(1/z))=0 ∴z_0 =−(e/(e−1)) inf[f(z),z=−(e/(e−1))]=lim_(x→z_0 ) f(z)=−(1/e) But, z∈Z^+ /{0} we don′t need inf[f(z)] when z is less than 0 inf[f(z),z=∞]=0 (becauselim_(z→∞) f(z)=0) sup is not Exist. (because lim_(z→0) f(z)=∞, f^((1)) (a)<0 ,a∈Z^+ /{0})](https://www.tinkutara.com/question/Q174392.png)
$$\mathrm{Let}'\mathrm{s}\:\mathrm{think}\:\mathrm{of}\:\mathrm{a}\:\mathrm{Sequence}\:\mathrm{A}_{{k}} \:\mathrm{Function}\:\mathrm{as}\:{f}\left({z}\right) \\ $$$$\mathrm{and}\:\mathrm{Find}\:\mathrm{Root}\:{f}^{\left(\mathrm{1}\right)} \left({z}\right)=\mathrm{0}\:\mathrm{for}\:\mathrm{find}\:\mathrm{inf}\:\left[{f}\left({z}\right),{z}_{\mathrm{0}} \right] \\ $$$$\mathrm{so}\:,\:{f}\left({z}\right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right)\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right) \\ $$$$\frac{\mathrm{d}{f}\left({z}\right)}{\mathrm{d}{z}}=−\left(\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right)−\frac{\mathrm{1}+\frac{\mathrm{1}}{{z}}}{\mathrm{1}+\frac{\mathrm{1}}{{z}}}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right)=\mathrm{0} \\ $$$$\therefore{z}_{\mathrm{0}} =−\frac{{e}}{{e}−\mathrm{1}} \\ $$$$\mathrm{inf}\left[{f}\left({z}\right),{z}=−\frac{{e}}{{e}−\mathrm{1}}\right]=\underset{{x}\rightarrow{z}_{\mathrm{0}} } {\mathrm{lim}}{f}\left({z}\right)=−\frac{\mathrm{1}}{{e}} \\ $$$$\mathrm{But},\:{z}\in\mathbb{Z}^{+} /\left\{\mathrm{0}\right\}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{inf}\left[{f}\left({z}\right)\right]\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{0} \\ $$$$\:\mathrm{inf}\left[{f}\left({z}\right),{z}=\infty\right]=\mathrm{0}\:\:\left(\mathrm{because}\underset{{z}\rightarrow\infty} {\mathrm{lim}}{f}\left({z}\right)=\mathrm{0}\right) \\ $$$$\mathrm{sup}\:\mathrm{is}\:\mathrm{not}\:\mathrm{Exist}.\:\:\left(\mathrm{because}\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({z}\right)=\infty,\:{f}^{\left(\mathrm{1}\right)} \left({a}\right)<\mathrm{0}\:,{a}\in\mathbb{Z}^{+} /\left\{\mathrm{0}\right\}\right) \\ $$
Commented by Best1 last updated on 31/Jul/22
$${well}\:{but}\:{see}\:{it}\:{again} \\ $$