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Question Number 136070 by Dwaipayan Shikari last updated on 18/Mar/21

((sin((√2)))/1^3 )+((sin(2(√2)))/2^3 )+((sin(3(√2)))/3^3 )+...=((π^b +1)/( a(√b)))−(π/b)  Find a−b

$$\frac{{sin}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}^{\mathrm{3}} }+\frac{{sin}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)}{\mathrm{3}^{\mathrm{3}} }+...=\frac{\pi^{{b}} +\mathrm{1}}{\:{a}\sqrt{{b}}}−\frac{\pi}{{b}} \\ $$$${Find}\:{a}−{b} \\ $$

Answered by mnjuly1970 last updated on 18/Mar/21

  Ω(x)=Σ_(n=1) ^∞ ((sin(nx))/n^3 )   −Im(ln(1−e^(ix) ))=Σ_(n=1) =∞((sin(nx))/n)     −Imln((1−cos(x))−isin(x))=Σ_(n=1) ^∞ ((sin(nx))/n)  −Im{ln(2sin((x/2))−itan^(−1) (((sin(x))/(1−cos(x))))=Σ_(n=1) ^∞ ((sin(nx))/(n=1))  (π/2)−(x/2)=Σ_(n=1) ^∞ ((sin(nx))/n)    (π/2)x−(x^2 /4)=−Σ_(n=1) ^∞ ((cos(nx))/n^2 )+C    x=π     (π^2 /4)=(π^2 /(12))+C⇒C=(π^2 /6)  (π^2 /6)+(x^2 /4)−(π/2)x=Σ_(n=1) ^∞ ((cos(nx))/n^2 )   (π^2 /6)x+(x^3 /(12))−(π/4)x^2 =Σ_(n=1) ^∞ ((sin(nx))/n^3 )  x=(√2)     Σ_(n=1) ^∞ ((sin(n(√2)))/n^3 )=(π^2 /6)(√2) +((√2)/6)−(π/2)                         Ω=((π^2 +1)/(3(√2)))−(π/2)                       hence ::  a−b=3−2=1         please check  my solution....

$$\:\:\Omega\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{3}} } \\ $$$$\:−{Im}\left({ln}\left(\mathrm{1}−{e}^{{ix}} \right)\right)=\underset{{n}=\mathrm{1}} {\sum}=\infty\frac{{sin}\left({nx}\right)}{{n}} \\ $$$$\:\:\:−{Imln}\left(\left(\mathrm{1}−{cos}\left({x}\right)\right)−{isin}\left({x}\right)\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}} \\ $$$$−{Im}\left\{{ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)−{itan}^{−\mathrm{1}} \left(\frac{{sin}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}=\mathrm{1}}\right.\right. \\ $$$$\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}} \\ $$$$\:\:\frac{\pi}{\mathrm{2}}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }+{C} \\ $$$$\:\:{x}=\pi \\ $$$$\:\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+{C}\Rightarrow{C}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi}{\mathrm{2}}{x}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} } \\ $$$$\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{12}}−\frac{\pi}{\mathrm{4}}{x}^{\mathrm{2}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{3}} } \\ $$$${x}=\sqrt{\mathrm{2}} \\ $$$$\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\sqrt{\mathrm{2}}\right)}{{n}^{\mathrm{3}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\sqrt{\mathrm{2}}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{6}}−\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\frac{\pi^{\mathrm{2}} +\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{2}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{hence}\:::\:\:{a}−{b}=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{please}\:{check}\:\:{my}\:{solution}.... \\ $$$$\:\:\:\:\: \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 18/Mar/21

Great sir!

$${Great}\:{sir}! \\ $$

Commented by mnjuly1970 last updated on 18/Mar/21

thank you for your nice questions...

$${thank}\:{you}\:{for}\:{your}\:{nice}\:{questions}... \\ $$

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