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Question Number 217088 by alcohol last updated on 28/Feb/25

$$showthat{\int }_n^{n+1}ln\left(t\right)dt⩽ln(n+\frac{1}{2})$$$$Given{u}_n=\frac{{\left(4n\right)}^{n}n!e^{-n}}{\left(2n\right)!},\mathrm{\forall }n⩾1$$$$prove,usingtheprecedingquestionthat$$$$u_{n}isdecreasingandconvergent$$

Answered by MrGaster last updated on 01/Mar/25

$$\mathrm{Prove}:{\int }_n^{n+1}\ln \left(t\right)dt\le \ln (n+\frac{1}{2})$$$${\int }_n^{n+1}\ln \left(t\right)dt={[t\ln \left(t\right)-t]}_n^{n+1}$$$$=((n+1)+\ln (n+1)-(n+1)-(n\ln \left(n\right)-n)$$$$=(n+1)\ln (n+1)-n\ln \left(n\right)-1$$$$\ln (n+1)\le \ln \left(n\right)+\frac{1}{n}⊢(n+1)\ln (n+1)\le (n+1)(\ln \left(n\right)+\frac{1}{n})$$$$=(n+1)\ln \left(n\right)+(n+1)\frac{1}{n}$$$$=(n+1)\ln \left(n\right)+1\frac{1}{n}$$$$\therefore (n+1)\ln (n+1)-n\ln \left(n\right)-1\le ((n+1)\ln \left(n\right)+1+\frac{1}{n})-n\ln \left(n\right)-1$$$$=\ln \left(n\right)+\frac{1}{n}$$$$\because \frac{1}{n}\le \ln (1+\frac{1}{2n})\Rightarrow \ln \left(n\right)+\frac{1}{n}\le \ln \left(n\right)+\ln (1+\frac{1}{2n})$$$$=\ln \left(n(1+\frac{1}{2n})\right)$$$$=\ln (n+\frac{1}{2})$$$$\mathrm{so},\overline{)\begin{array}{c}{\int }_n^{n+1}\ln \left(t\right)dt\le \ln (n+\frac{1}{2})\end{array}}$$$$\mathrm{Give}{u}_n=\frac{{\left(4n\right)}^{n}n!e^{-n}}{\left(2n\right)!},\mathrm{\forall }n\ge 1,\mathrm{Prove}\mathrm{that}{u}_n\mathrm{is}\mathrm{decreasing}\mathrm{and}\mathrm{converhgent}.$$$$\mathrm{Part}1:\mathrm{Prove}{u}_n\mathrm{is}\mathrm{decreasing}.$$$$\mathrm{Consider}\mathrm{the}\mathrm{ratio}:$$$$\frac{u_{n+1}}{u_n}=\frac{\frac{(4{(n+1)}^{n+1}(n+1)!e^{-(n+1)}}{(2(n+1)!}}{\frac{{\left(4n\right)}^n{e}^{-n}}{\left(2n\right)!}}$$$$=\frac{{\left(4(n+1)\right)}^{n+1}(n+1)e^{-1}}{{\left(4n\right)}^n(2n+2)(2n+1)}$$$$=\frac{4^{n+1}{(n+1)}^{n+1}(n+1)e^{-1}}{4^n{n}^n(2n+2)(2n+1)}$$$$=\frac{4{(n+1)}^{n+2}{e}^{-1}}{n^{n}2(n+1)(2n+1)}$$$$=\frac{2{(n+1)}^{n+1}{e}^{-1}}{n^n(2n+1)}$$$$\because {(n+1)}^{n+1}>n^{n+1}\Rightarrow \frac{{(n+1)}^{n-1}}{n^n}>1$$$$\therefore \frac{u_{n+1}}{u_n}<1\mathrm{for}\mathrm{large}n,\mathrm{implying}{u}_n\mathrm{is}\mathrm{decreasing}.$$$$\mathrm{Step}2:\mathrm{Prove}{u}_n\mathrm{is}\mathrm{convergent}.\mathrm{Using}{\mathrm{stirling}}^{\prime }\mathrm{s}\mathrm{approximation}n!\approx \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n\Rightarrow u_n=\frac{{\left(4n\right)}^{n}n!e^{-n}}{\left(2n\right)!}\approx \frac{{\left(4n\right)}^n\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n{e}^{-n}}{\sqrt{4\pi n}{\left(\frac{2n}{e}\right)}^{2n}}$$$$=\frac{{\left(4n\right)}^n\sqrt{2\pi n}{n}^n{e}^{-2n}}{\sqrt{4\pi n}{2}^{2n}{n}^{2n}}$$$$=\frac{\sqrt{2\pi n}{4}^n{n}^n{e}^{-2n}}{\sqrt{4\pi n}{2}^n{n}^{2n}}$$$$=\frac{\sqrt{2\pi }{4}^n{e}^{-2n}}{\sqrt{4\pi }{2}^{2n}{n}^n}$$$$=\frac{\sqrt{2}{2}^{2n}{e}^{-2n}}{\sqrt{2}{2}^{2n}{n}^n}$$$$=\frac{e^{2n}}{n^n}$$$$\mathrm{As}n\to \mathrm{\infty },e^{-2n}\to 0\mathrm{and}{n}^n\to \mathrm{\infty },\mathrm{so}{u}_n\to 0.\mathrm{Thus},u_n\mathrm{is}\mathrm{convergent}.$$$$\overline{)\begin{array}{c}{u}_n\mathrm{is}\mathrm{decreasing}\mathrm{and}\mathrm{convergent}\end{array}}$$

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