Question Number 123764 by mnjuly1970 last updated on 28/Nov/20
$$\:\:\:\:\:\:\:\:\:\:\:\:….\:{advanced}\:\:{calculus}\:… \\ $$$$\:\:\:\:\:\:\:{prove}\:\:{that}:::: \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\zeta\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{4}^{{n}\:} \:\left(\mathrm{2}{n}+\mathrm{1}\right)}\right\}={ln}\left(\mathrm{2}\right)−\gamma \\ $$$$\:\:\:\:\:\:\:\:\gamma::\:\:{euler}−{mascheroni} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{constant} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Nov/20
![Σ_(n=1) ^∞ ((ζ(2n+1))/(4^n (2n+1))) Σ_(n=1) ^∞ Σ_(k=1) ^∞ (1/(4^n (2n+1)k^((2n+1)) )) Σ_(k=1) ^∞ Σ_(n≥1) ^∞ (1/(k(4k^2 )^n (2n+1))) Σ_(k≥1) ^∞ Σ_(n≥1) ^∞ ∫_0 ^1 (x^(2n) /(k(4k^2 )^n ))dx Σ_(k≥1) ^∞ ∫_0 ^1 Σ_(n≥1) ^∞ (x^(2n) /(k(4k^2 )^n ))dx=Σ_(k≥1) ^∞ ∫_0 ^1 (1/k).((x^2 /(4k^2 ))/(1−(x^2 /(4k^2 )))) dx =Σ_(k≥1) ^∞ ∫_0 ^1 (1/k).(x^2 /(4k^2 −x^2 ))dx (1/2)∫_0 ^1 x(Σ_(k≥1) ^∞ (1/k).(1/((2k−x)))−(1/k).(1/((2k+x)))) dx =−(1/2)(∫_0 ^1 Σ_(k≥1) ^∞ (1/k)−(1/(2k−x))−∫_0 ^1 (1/(2k))−(1/(2k+x))) =−(1/2)((∫_0 ^1 Σ^∞ (1/k)−(1/(k−(x/2)))+Σ^∞ (1/k)−(1/(k+(x/2))) ))dx =−γ−(1/2)∫_0 ^1 ψ(−(x/2))+ψ((x/2)) =−γ−[log(Γ((x/2))−log(Γ(−(x/2)))]_0 ^1 =−γ+log(2) =log(2)−γ](https://www.tinkutara.com/question/Q123775.png)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right){k}^{\left(\mathrm{2}{n}+\mathrm{1}\right)} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}\left(\mathrm{4}{k}^{\mathrm{2}} \right)^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}} }{{k}\left(\mathrm{4}{k}^{\mathrm{2}} \right)^{{n}} }{dx} \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{{k}\left(\mathrm{4}{k}^{\mathrm{2}} \right)^{{n}} }{dx}=\underset{{k}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{k}}.\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{4}{k}^{\mathrm{2}} }}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}{k}^{\mathrm{2}} }}\:{dx} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{k}}.\frac{{x}^{\mathrm{2}} }{\mathrm{4}{k}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\underset{{k}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}.\frac{\mathrm{1}}{\left(\mathrm{2}{k}−{x}\right)}−\frac{\mathrm{1}}{{k}}.\frac{\mathrm{1}}{\left(\mathrm{2}{k}+{x}\right)}\right)\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\mathrm{2}{k}−{x}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{k}}−\frac{\mathrm{1}}{\mathrm{2}{k}+{x}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\int_{\mathrm{0}} ^{\mathrm{1}} \overset{\infty} {\sum}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}−\frac{{x}}{\mathrm{2}}}+\overset{\infty} {\sum}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\frac{{x}}{\mathrm{2}}}\:\right)\right){dx} \\ $$$$=−\gamma−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \psi\left(−\frac{{x}}{\mathrm{2}}\right)+\psi\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$=−\gamma−\left[{log}\left(\Gamma\left(\frac{{x}}{\mathrm{2}}\right)−{log}\left(\Gamma\left(−\frac{{x}}{\mathrm{2}}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \right. \\ $$$$=−\gamma+{log}\left(\mathrm{2}\right) \\ $$$$={log}\left(\mathrm{2}\right)−\gamma \\ $$
Commented by mnjuly1970 last updated on 28/Nov/20
$${God}\:{keep}\:{you} \\ $$$${excellent}\:{mr}\:{payan}… \\ $$