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Question-128767

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Question Number 128767 by LUFFY last updated on 10/Jan/21
Commented by LUFFY last updated on 10/Jan/21
ans (1/(24)) but i am getting (1/(12))
$$\mathrm{ans}\:\frac{\mathrm{1}}{\mathrm{24}}\:\mathrm{but}\:\mathrm{i}\:\mathrm{am}\:\mathrm{getting}\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$
Commented by LUFFY last updated on 10/Jan/21
send me please
$$\boldsymbol{{send}}\:\boldsymbol{{me}}\:\boldsymbol{{please}} \\ $$
Answered by Dwaipayan Shikari last updated on 10/Jan/21
Σ_(n=1) ^∞ (n^k /(e^(2πn) −1))=((k!)/((2π)^(k+1) )).ζ(k+1) (k=4m+1) k=13 It is ((13!)/((2π)^(14) )).(((2π)^(14) B_(14) )/(2(14)!))=(B_(14) /(28))=(1/(24)) B_n =Bernoulli Number
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{k}} }{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}}=\frac{{k}!}{\left(\mathrm{2}\pi\right)^{{k}+\mathrm{1}} }.\zeta\left({k}+\mathrm{1}\right)\:\:\:\:\:\:\left({k}=\mathrm{4}{m}+\mathrm{1}\right) \\ $$$${k}=\mathrm{13}\:\:\:\:\:{It}\:{is}\:\frac{\mathrm{13}!}{\left(\mathrm{2}\pi\right)^{\mathrm{14}} }.\frac{\left(\mathrm{2}\pi\right)^{\mathrm{14}} {B}_{\mathrm{14}} }{\mathrm{2}\left(\mathrm{14}\right)!}=\frac{{B}_{\mathrm{14}} }{\mathrm{28}}=\frac{\mathrm{1}}{\mathrm{24}} \\ $$$$\boldsymbol{\mathrm{B}}_{{n}} ={Bernoulli}\:{Number} \\ $$
Commented by LUFFY last updated on 10/Jan/21
check again
$$\mathrm{check}\:\mathrm{again} \\ $$
Commented by LUFFY last updated on 10/Jan/21
can you send full solution
$$\mathrm{can}\:\mathrm{you}\:\mathrm{send}\:\mathrm{full}\:\mathrm{solution} \\ $$
Commented by Dwaipayan Shikari last updated on 19/Feb/21
(1/(e^(2π) −1))=Σ_(n=1) ^∞ e^(−2πn ) Σ_(n=1) ^∞ (1/(n^2 +x^2 ))=(π/(2x))coth(πx)−(1/(2x^2 )) (1/x)+2xΣ_(n=1) ^∞ (1/(n^2 +x^2 ))=πcoth(πx)⇒(1/(2x))+xΣ_(n=1) ^∞ (1/(n^2 +x^2 ))=(π/2)+(π/(e^(2πx) −1)) (1/(e^(2πx) −1))=(x/π)Σ_(n=1) ^∞ (1/(n^2 +x^2 ))+(1/(2πx))−(1/2) Σ_(n=1) ^∞ e^(−2πnx) =(1/(2πx))Σ_(n=1) ^∞ (1/(x+in))+(1/(x−in))+(1/(2πx))−(1/2) differentiating k th time (k=4m+1) (2π)^k Σ_(x=1) ^∞ n^k e^(−2πnx) =((k!)/((x)^(k+1) (2π)))+Σ_(n=1) ^∞ (1/((x+in)^(k+1) ))+(1/((x−in)^(k+1) )) Σ_(n=1) ^∞ (n^k /(e^(2πn) −1))=((k!)/((2π)^(k+1) ))Σ_(x=1) ^∞ (1/x^(k+1) )+Σ_(n=1) ^∞ Σ_(x=1) ^∞ (1/((x+in)^(k+1) ))+(1/((x−in)^(k+1) )) ⇒Σ_(n=1) ^∞ (n^k /(e^(2πn) −1))=((k!)/((2π)^(k+1) ))ζ(k+1) (k=4m+1) Σ_(x≥1) Σ_(n≥1) (1/((x+in)^(k+1) ))+(1/((x−in)^(k+1) ))=0
$$\frac{\mathrm{1}}{{e}^{\mathrm{2}\pi} −\mathrm{1}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}\pi{n}\:\:} \:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{x}}{coth}\left(\pi{x}\right)−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}}+\mathrm{2}{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\pi{coth}\left(\pi{x}\right)\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{x}}+{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}+\frac{\pi}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}=\frac{{x}}{\pi}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}\pi{x}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}\pi{nx}} =\frac{\mathrm{1}}{\mathrm{2}\pi{x}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}+{in}}+\frac{\mathrm{1}}{{x}−{in}}+\frac{\mathrm{1}}{\mathrm{2}\pi{x}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${differentiating}\:\:\:{k}\:{th}\:{time}\:\:\left({k}=\mathrm{4}{m}+\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\pi\right)^{{k}} \underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{{k}} {e}^{−\mathrm{2}\pi{nx}} =\frac{{k}!}{\left({x}\right)^{{k}+\mathrm{1}} \left(\mathrm{2}\pi\right)}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({x}+{in}\right)^{{k}+\mathrm{1}} }+\frac{\mathrm{1}}{\left({x}−{in}\right)^{{k}+\mathrm{1}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{k}} }{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}}=\frac{{k}!}{\left(\mathrm{2}\pi\right)^{{k}+\mathrm{1}} }\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}^{{k}+\mathrm{1}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({x}+{in}\right)^{{k}+\mathrm{1}} }+\frac{\mathrm{1}}{\left({x}−{in}\right)^{{k}+\mathrm{1}} } \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{k}} }{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}}=\frac{{k}!}{\left(\mathrm{2}\pi\right)^{{k}+\mathrm{1}} }\zeta\left({k}+\mathrm{1}\right)\:\:\:\:\left({k}=\mathrm{4}{m}+\mathrm{1}\right) \\ $$$$\underset{{x}\geqslant\mathrm{1}} {\sum}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({x}+{in}\right)^{{k}+\mathrm{1}} }+\frac{\mathrm{1}}{\left({x}−{in}\right)^{{k}+\mathrm{1}} }=\mathrm{0} \\ $$
Commented by Dwaipayan Shikari last updated on 10/Jan/21
For every k=4m+1 , this is valid Σ_(n=1) ^∞ (n^5 /(e^(2πn) −1))=(1/(504))...
$${For}\:{every}\:{k}=\mathrm{4}{m}+\mathrm{1}\:,\:{this}\:{is}\:{valid}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{5}} }{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{504}}… \\ $$

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