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Question Number 195733 by York12 last updated on 09/Aug/23
prove that Σ_(n=2) ^∞ [(B_n^_ /((n−2)!))]=((e(3−e))/((e−1)^3 )) where B_n^_ is the n− th bernouli′s number
$${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left[\frac{{B}_{\overset{\_} {{n}}} }{\left({n}−\mathrm{2}\right)!}\right]=\frac{{e}\left(\mathrm{3}−{e}\right)}{\left({e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${where}\:{B}_{\overset{\_} {{n}}} \:{is}\:{the}\:{n}−\:{th}\:{bernouli}'{s}\:{number} \\ $$
Answered by witcher3 last updated on 09/Aug/23
Σ_(n≥0) B_n (x^n /(n!))=(x/(e^x −1))=f(x) f′(x)=Σ_(n≥1) (x^(n−1) /((n−1)!))B_n f′′(x)=Σ_(n≥2) B_n (x^(n−2) /((n−2)!)) f′(x)=(((e^x −1)−xe^x )/((e^x −1)^2 )) f′′(x)=((−xe^x (e^x −1)^2 −2e^x (e^x −1)(e^x −1−xe^x ))/((e^x −1)^4 )) =((−xe^(2x) +xe^x −2e^(2x) +2e^x +2xe^(2x) )/((e^x −1)^3 )) =((xe^(2x) +xe^x −2e^(2x) +2e^x )/((e^x −1)^3 )) f′′(1)=((3e−e^2 )/((e−1)^3 ))=Σ_(n≥1) (B_n /((n−2)!))
$$\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{B}_{\mathrm{n}} \frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\frac{\mathrm{x}}{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{x}^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{n}−\mathrm{1}\right)!}\mathrm{B}_{\mathrm{n}} \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)=\underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\mathrm{B}_{\mathrm{n}} \frac{\mathrm{x}^{\mathrm{n}−\mathrm{2}} }{\left(\mathrm{n}−\mathrm{2}\right)!} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)−\mathrm{xe}^{\mathrm{x}} }{\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)=\frac{−\mathrm{xe}^{\mathrm{x}} \left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2e}^{\mathrm{x}} \left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}−\mathrm{xe}^{\mathrm{x}} \right)}{\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{−\mathrm{xe}^{\mathrm{2x}} +\mathrm{xe}^{\mathrm{x}} −\mathrm{2e}^{\mathrm{2x}} +\mathrm{2e}^{\mathrm{x}} +\mathrm{2xe}^{\mathrm{2x}} }{\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{xe}^{\mathrm{2x}} +\mathrm{xe}^{\mathrm{x}} −\mathrm{2e}^{\mathrm{2x}} +\mathrm{2e}^{\mathrm{x}} }{\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{f}''\left(\mathrm{1}\right)=\frac{\mathrm{3e}−\mathrm{e}^{\mathrm{2}} }{\left(\mathrm{e}−\mathrm{1}\right)^{\mathrm{3}} }=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{B}_{\mathrm{n}} }{\left(\mathrm{n}−\mathrm{2}\right)!} \\ $$$$ \\ $$
Commented by York12 last updated on 10/Aug/23
I can not find words , but thanks so much sir
$${I}\:{can}\:{not}\:{find}\:{words}\:,\:{but}\:{thanks}\:{so}\:{much} \\ $$$${sir} \\ $$

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