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Question Number 58827 by jimful last updated on 30/Apr/19

$$\mathrm{Find}\sum _{\mathrm{x}=1}^{\mathrm{\infty }}(\frac{1}{\mathrm{x}}-\sin \left(\frac{1}{\mathrm{x}}\right))$$$$$$

Commented by tanmay last updated on 30/Apr/19

$$takingthehelpofgraphwecanfindthevalue$$

Commented by jimful last updated on 01/May/19

$$\mathrm{thanks}.\mathrm{I}\mathrm{need}\mathrm{to}\mathrm{learn}\mathrm{more}$$

Commented by maxmathsup by imad last updated on 30/Apr/19

$$wehavesinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\Rightarrow$$$$x-\frac{x^3}{6}⩽sinx⩽x\Rightarrow \frac{1}{x}-\frac{1}{6x^3}⩽sin\left(\frac{1}{x}\right)⩽\frac{1}{x}$$$$\Rightarrow -\frac{1}{x}⩽-sin\left(\frac{1}{x}\right)⩽-\frac{1}{x}+\frac{1}{6x^3}\Rightarrow 0⩽\frac{1}{x}-sin\left(\frac{1}{x}\right)⩽\frac{1}{6x^3}\Rightarrow$$$$0⩽\sum _{x=1}^{\mathrm{\infty }}(\frac{1}{x}-sin\left(\frac{1}{x}\right))⩽\frac{1}{6}\sum _{x=1}^{\mathrm{\infty }}\frac{1}{x^3}\Rightarrow 0⩽S⩽\frac{1}{6}\xi \left(3\right)...$$

Answered by tanmay last updated on 30/Apr/19

$$f\left(x\right)=\frac{1}{x}f\left(1\right)=1$$$$g\left(x\right)=sin\left(\frac{1}{x}\right)g\left(1\right)=sin\left(1\right)=0.841$$$$f\left(1\right)-g\left(1\right)=1-0.841=0.159$$$$$$$$f\left(2\right)=\frac{1}{2}=0.5$$$$g\left(2\right)=sin\left(\frac{1}{2}\right)=0.479$$$$f\left(2\right)-g\left(2\right)=0.5-0.479=0.021$$$$$$$$△=f\left(x\right)-g\left(x\right)$$$$thusweseeasweincreasethevalueofx$$$$△\to 0$$$$so\underset{x=1}{\sum ^{\mathrm{\infty }}}\frac{1}{x}-sin\left(\frac{1}{x}\right)=0.159+0.021+ϵ$$$$\underset{x=1}{\sum ^{\mathrm{\infty }}}\frac{1}{x}-sin\left(\frac{1}{x}\right)=0.18+ϵ$$$$ϵ=smallpositivenumber$$$$ihavetriedtofind..itisnotexactsolution$$$$butinterpretationoftheproblem...$$$$$$

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