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Question Number 122185 by mnjuly1970 last updated on 14/Nov/20

... challanging math... prove that :: Σ_(n=2) ^∞ (((−1)^n ζ(n))/(2^(n−1) n))=^(???) γ +ln((π/4)) hint :: ζ(s)=(1/(Γ(s)))∫_0 ^( ∞) (x^(s−1) /(e^x −1))dx m.n...

$$\:\:\:\:\:\:...\:{challanging}\:\:{math}... \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}−\mathrm{1}} {n}}\overset{???} {=}\gamma\:+{ln}\left(\frac{\pi}{\mathrm{4}}\right)\: \\ $$$$\:\:{hint}\:::\:\zeta\left({s}\right)=\frac{\mathrm{1}}{\Gamma\left({s}\right)}\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{s}−\mathrm{1}} }{{e}^{{x}} −\mathrm{1}}{dx} \\ $$$$\:\:\:\:\:\:\:{m}.{n}... \\ $$

Commented by Dwaipayan Shikari last updated on 15/Nov/20

Σ_(n=2) ^∞ (((−1)^n ζ(n))/(2^(n−1) n)) Σ_(n=2) ^∞ (((−1)^n )/(2^(n−1) n))Σ_(s=1) ^∞ (1/s^n ) 2Σ_(n=2) ^∞ Σ_(s=1) ^∞ (((−1)^n )/((2s)^n n)) 2Σ_(s=1) ^∞ (((((−1)/(2s)))^n )/n)=2Σ_(s=1) ^∞ log(1+(1/(2s)))+(1/(2s)) Σ_(s=1) ^∞ (1/s)+2Σ^∞ log(1+(1/(2s))) log(Γ((1/2)))=−(γ/2)+log2+Σ^∞ (1/(2s))+Σ^∞ log(1+(1/(2s))) log(π)=−γ+log4+Σ^∞ (1/s)+2Σ^∞ log(1+(1/(2s))) Σ^∞ (1/s)+2Σ^∞ log(1+(1/(2s)))=γ+log((π/4))

$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}−\mathrm{1}} {n}} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}−\mathrm{1}} {n}}\underset{{s}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}^{{n}} } \\ $$$$\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\underset{{s}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{s}\right)^{{n}} {n}} \\ $$$$\mathrm{2}\underset{{s}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\frac{−\mathrm{1}}{\mathrm{2}{s}}\right)^{{n}} }{{n}}=\mathrm{2}\underset{{s}=\mathrm{1}} {\overset{\infty} {\sum}}{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{s}}\right)+\frac{\mathrm{1}}{\mathrm{2}{s}} \\ $$$$\underset{{s}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}}+\mathrm{2}\overset{\infty} {\sum}{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{s}}\right) \\ $$$${log}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)=−\frac{\gamma}{\mathrm{2}}+{log}\mathrm{2}+\overset{\infty} {\sum}\frac{\mathrm{1}}{\mathrm{2}{s}}+\overset{\infty} {\sum}{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{s}}\right) \\ $$$${log}\left(\pi\right)=−\gamma+{log}\mathrm{4}+\overset{\infty} {\sum}\frac{\mathrm{1}}{{s}}+\mathrm{2}\overset{\infty} {\sum}{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{s}}\right) \\ $$$$\overset{\infty} {\sum}\frac{\mathrm{1}}{{s}}+\mathrm{2}\overset{\infty} {\sum}{log}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{s}}\right)=\gamma+{log}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$

Commented by mnjuly1970 last updated on 15/Nov/20

very nice .bravo master dwaipayan..

$${very}\:{nice}\:.{bravo} \\ $$$$\:\:{master}\:\:{dwaipayan}.. \\ $$

Answered by mindispower last updated on 15/Nov/20

Σ_(n≥2) (((−1)^n x^n ζ(n))/n) =Σ_(n≥2) (((−x)^n )/n).Σ_(k≥1) (1/k^n ) =Σ_(k≥1) Σ_(n≥2) (((−(x/k))^n )/n)=S(x) =Σ_(k≥1) −(ln(1+(x/k))−(x/k))=S(x) Γ(x)=(e^(−γx) /x)Π_(k≥1) (e^(x/k) /(1+(x/k)))⇒ log(Γ(z))−γx−ln(x)+ Σ_(k≥1) ((x/k)−ln(1+(x/k))) ⇒S(x)=log(Γ(x))+γx+ln(x) =S(x)=log(xΓ(x))+γx=log(Γ(x+1))+γx Σ(((−1)^n ζ(n))/(2^(n−1) n))=2Σ_(n≥2) (((−1)^n ((1/2))^n ζ(n))/n)=2S((1/2)) 2(log(Γ((1/2))+(γ/2))=log(π/4)+γ withe hint leads to ∫_0 ^∞ (2/x)(e^(−(x/2)) −1+(x/2)).(dx/(e^x −1))

$$\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \zeta\left({n}\right)}{{n}} \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−{x}\right)^{{n}} }{{n}}.\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}^{{n}} } \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\frac{{x}}{{k}}\right)^{{n}} }{{n}}={S}\left({x}\right) \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}−\left({ln}\left(\mathrm{1}+\frac{{x}}{{k}}\right)−\frac{{x}}{{k}}\right)={S}\left({x}\right) \\ $$$$\Gamma\left({x}\right)=\frac{{e}^{−\gamma{x}} }{{x}}\underset{{k}\geqslant\mathrm{1}} {\prod}\frac{{e}^{\frac{{x}}{{k}}} }{\mathrm{1}+\frac{{x}}{{k}}}\Rightarrow \\ $$$${log}\left(\Gamma\left({z}\right)\right)−\gamma{x}−{ln}\left({x}\right)+\:\underset{{k}\geqslant\mathrm{1}} {\sum}\left(\frac{{x}}{{k}}−{ln}\left(\mathrm{1}+\frac{{x}}{{k}}\right)\right) \\ $$$$\Rightarrow{S}\left({x}\right)={log}\left(\Gamma\left({x}\right)\right)+\gamma{x}+{ln}\left({x}\right) \\ $$$$={S}\left({x}\right)={log}\left({x}\Gamma\left({x}\right)\right)+\gamma{x}={log}\left(\Gamma\left({x}+\mathrm{1}\right)\right)+\gamma{x} \\ $$$$\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}−\mathrm{1}} {n}}=\mathrm{2}\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \zeta\left({n}\right)}{{n}}=\mathrm{2}{S}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{2}\left({log}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\gamma}{\mathrm{2}}\right)={log}\frac{\pi}{\mathrm{4}}+\gamma\right. \\ $$$${withe}\:{hint}\:{leads}\:{to} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}}{{x}}\left({e}^{−\frac{{x}}{\mathrm{2}}} −\mathrm{1}+\frac{{x}}{\mathrm{2}}\right).\frac{{dx}}{{e}^{{x}} −\mathrm{1}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 15/Nov/20

peace be upon you sir mindspower.really nice and excellent.

$${peace}\:{be}\:{upon}\:{you}\:{sir}\: \\ $$$${mindspower}.{really}\:{nice}\:{and} \\ $$$${excellent}. \\ $$

Answered by mnjuly1970 last updated on 15/Nov/20

Answered by mnjuly1970 last updated on 15/Nov/20

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