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Question Number 74554 by shubham90412@gmail.com last updated on 26/Nov/19

$$\mathit{F}\mathit{i}\mathit{n}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{u}\mathit{p}\mathit{e}\mathit{r}\mathit{i}\mathit{m}\mathit{u}\mathit{m}\mathit{o}\mathit{f}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{e}\mathit{t}\left\{\frac{{\mathit{n}}^2}{2^{\mathit{n}}}\right\}$$

Answered by MJS last updated on 26/Nov/19

$$\mathrm{trying}$$$$S=\{0,\frac{1}{2},1,\frac{9}{8},1,\frac{25}{32},\frac{9}{16},...\}$$$$$$$$\mathrm{calculating}$$$$f\left(x\right)=\frac{x^2}{2^x}$$$$f^{\prime }\left(x\right)=\frac{x}{2^x}(2-x\ln 2)$$$$\frac{x}{2^x}(2-x\ln 2)=0\Rightarrow x=0\vee x=\frac{2}{\ln 2}\approx 2.89$$$$f^{″}\left(x\right)=\frac{{\left(\ln 2\right)}^2{x}^2-4x\ln 2+2}{2^x}$$$$f^{″}\left(0\right)>0\Rightarrow x=0\mathrm{is}\mathrm{minimum}$$$$f^{″}\left(\frac{2}{\ln 2}\right)<0\Rightarrow x=\frac{2}{\ln 2}\mathrm{is}\mathrm{maximum}$$$$\Rightarrow \mathrm{we}\mathrm{have}\mathrm{to}\mathrm{try}n=\lfloor \frac{2}{\ln 2}\rfloor =2\vee n=\lceil \frac{2}{\ln 2}\rceil =3$$$$\Rightarrow \mathrm{supremium}\mathrm{at}n=3$$

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