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

Find Σ_(x=1) ^∞ ((1/x)−sin((1/x)))

$$\mathrm{Find}\:\:\sum_{\mathrm{x}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{x}}−\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right) \\ $$$$ \\ $$

Commented by tanmay last updated on 30/Apr/19

taking the help of graph we can find the value

$${taking}\:{the}\:{help}\:{of}\:{graph}\:{we}\:{can}\:{find}\:{the}\:{value} \\ $$

Commented by jimful last updated on 01/May/19

thanks. I need to learn more

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

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

we have sinx =Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)!)) =x−(x^3 /(3!)) +(x^5 /(5!)) −... ⇒ x−(x^3 /6) ≤ sinx ≤x ⇒ (1/x) −(1/(6x^3 )) ≤ sin((1/x))≤(1/x) ⇒ −(1/x) ≤−sin((1/x))≤−(1/x) +(1/(6x^3 )) ⇒0≤ (1/x) −sin((1/x)) ≤ (1/(6x^3 )) ⇒ 0≤ Σ_(x=1) ^∞ ((1/x) −sin((1/x))) ≤(1/6) Σ_(x=1) ^∞ (1/x^3 ) ⇒ 0≤ S ≤(1/6)ξ(3) ...

$${we}\:{have}\:{sinx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}\:−...\:\Rightarrow \\ $$$${x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant\:{sinx}\:\leqslant{x}\:\Rightarrow\:\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{\mathrm{6}{x}^{\mathrm{3}} }\:\leqslant\:{sin}\left(\frac{\mathrm{1}}{{x}}\right)\leqslant\frac{\mathrm{1}}{{x}} \\ $$$$\:\Rightarrow\:−\frac{\mathrm{1}}{{x}}\:\leqslant−{sin}\left(\frac{\mathrm{1}}{{x}}\right)\leqslant−\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{6}{x}^{\mathrm{3}} }\:\Rightarrow\mathrm{0}\leqslant\:\frac{\mathrm{1}}{{x}}\:−{sin}\left(\frac{\mathrm{1}}{{x}}\right)\:\leqslant\:\frac{\mathrm{1}}{\mathrm{6}{x}^{\mathrm{3}} }\:\Rightarrow \\ $$$$\mathrm{0}\leqslant\:\sum_{{x}=\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{{x}}\:−{sin}\left(\frac{\mathrm{1}}{{x}}\right)\right)\:\leqslant\frac{\mathrm{1}}{\mathrm{6}}\:\sum_{{x}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:\Rightarrow\:\mathrm{0}\leqslant\:{S}\:\leqslant\frac{\mathrm{1}}{\mathrm{6}}\xi\left(\mathrm{3}\right)\:... \\ $$

Answered by tanmay last updated on 30/Apr/19

f(x)=(1/x) f(1)=1 g(x)=sin((1/x)) g(1)=sin(1)=0.841 f(1)−g(1)=1−0.841=0.159 f(2)=(1/2)=0.5 g(2)=sin((1/2))=0.479 f(2)−g(2)=0.5−0.479=0.021 △=f(x)−g(x) thus we see as we increase the value of x △→0 so Σ_(x=1) ^∞ (1/x)−sin((1/x))=0.159+0.021+ε Σ_(x=1) ^∞ (1/x)−sin((1/x))=0.18+ε ε=small positive number i have tried to find ..it is not exact solution but interpretation of the problem...

$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${g}\left({x}\right)={sin}\left(\frac{\mathrm{1}}{{x}}\right)\:\:\:\:\:\:{g}\left(\mathrm{1}\right)={sin}\left(\mathrm{1}\right)=\mathrm{0}.\mathrm{841} \\ $$$${f}\left(\mathrm{1}\right)−{g}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{0}.\mathrm{841}=\mathrm{0}.\mathrm{159} \\ $$$$ \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}.\mathrm{5} \\ $$$${g}\left(\mathrm{2}\right)={sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}.\mathrm{479} \\ $$$${f}\left(\mathrm{2}\right)−{g}\left(\mathrm{2}\right)=\mathrm{0}.\mathrm{5}−\mathrm{0}.\mathrm{479}=\mathrm{0}.\mathrm{021} \\ $$$$ \\ $$$$\bigtriangleup={f}\left({x}\right)−{g}\left({x}\right) \\ $$$${thus}\:{we}\:{see}\:{as}\:{we}\:{increase}\:{the}\:{value}\:{of}\:{x}\: \\ $$$$\bigtriangleup\rightarrow\mathrm{0} \\ $$$${so}\:\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}−{sin}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{0}.\mathrm{159}+\mathrm{0}.\mathrm{021}+\epsilon \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{x}}−{sin}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{0}.\mathrm{18}+\epsilon \\ $$$$\epsilon={small}\:{positive}\:{number} \\ $$$${i}\:{have}\:{tried}\:{to}\:{find}\:..{it}\:{is}\:{not}\:{exact}\:{solution} \\ $$$${but}\:{interpretation}\:{of}\:{the}\:{problem}... \\ $$$$ \\ $$

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