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Question Number 178777 by mr W last updated on 21/Oct/22

$$A=\frac{1}{2018}+\frac{1}{2019}+...+\frac{1}{2050}$$$$findtheintegerpartof\frac{1}{A}.$$

Commented by Frix last updated on 21/Oct/22

$$\frac{1}{A}\mathrm{should}\mathrm{be}\mathrm{close}\mathrm{to}\frac{2018+2050}{2\times 33}=61.6363...$$$$\Rightarrow \mathrm{answer}\mathrm{should}\mathrm{be}61$$

Commented by mr W last updated on 21/Oct/22

$$howis\frac{1}{A}closeto\frac{2018+2050}{2\times 33}?$$$$weneedstrictproof.$$

Commented by Mathematification last updated on 21/Oct/22

$$\mathrm{Workings}\mathrm{please}$$

Commented by Frix last updated on 22/Oct/22

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{an}\mathrm{idea}...$$$$A=\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{n+k-1}$$$$\Rightarrow$$$$\frac{k}{A}\mathrm{is}\mathrm{the}\mathrm{harmonic}\mathrm{mean}\mathrm{of}\{n,n+1,...,n+k-1\}$$$$\mathrm{if}q=\frac{n+k-1}{n}\mathrm{is}\mathrm{close}\mathrm{to}1\iff n\mathrm{is}\mathrm{big}\mathrm{enough}\mathrm{and}$$$$k\mathrm{is}\mathrm{not}\mathrm{too}\mathrm{big}\frac{k}{A}\mathrm{is}\mathrm{very}\mathrm{close}\mathrm{to}\frac{n+(n+k-1)}{2}$$$$\mathrm{I}\mathrm{cannot}\mathrm{prove}\mathrm{this}.\mathrm{just}\mathrm{tried}\mathrm{these}:$$$$n=10;k=10;q=1.9;\frac{k}{A}\approx 13.91;\frac{10+19}{2}=14.5good$$$$n=100;k=10;q=1.09;\frac{k}{A}\approx 104.42;\frac{100+109}{2}=104.5better$$$$n=10;k=100;q=10.9;\frac{k}{A}\approx 40.91;\frac{10+109}{2}=59.5bad$$$$\mathrm{so}\mathrm{I}\mathrm{guessed}\mathrm{that}\mathrm{with}$$$$n=2018;k=33;q=1.016{\mathrm{we}}^{\prime }\mathrm{d}\mathrm{get}\frac{k}{A}\mathrm{close}$$$$\mathrm{to}\frac{2018+2050}{2}=2034\mathrm{and}\frac{1}{A}\mathrm{close}\mathrm{to}\frac{2034}{33}$$$$\mathrm{which}\mathrm{is}\approx 61.64$$

Commented by mr W last updated on 22/Oct/22

$$thanksfortheexplanationsir!$$

Commented by Tawa11 last updated on 22/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 22/Oct/22

$$A=\underset{18numbers}{(\frac{1}{2018}+\frac{1}{2019}+...+\frac{1}{2035})}+\underset{15numbers}{(\frac{1}{2036}+...+\frac{1}{2050})}$$$$A>\frac{18}{2035}+\frac{15}{2050}=\frac{15\times 2035+18\times 2050}{2050\times 2035}$$$$\Rightarrow \frac{1}{A}<\frac{2050\times 2035}{15\times 2035+18\times 2050}=\frac{166870}{2697}$$$$<\frac{167214}{2697}=62$$$$$$$$A=\underset{33numbers}{\frac{1}{2018}+\frac{1}{2019}+...+\frac{1}{2050}}$$$$A<\frac{33}{2018}$$$$\Rightarrow \frac{1}{A}>\frac{2018}{33}>\frac{2013}{33}=61$$$$$$$$since61<\frac{1}{A}<62,theintegerpartof$$$$\frac{1}{A}is61.$$

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