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Question Number 195733 by York12 last updated on 09/Aug/23

$$provethat$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\left[\frac{B_{\stackrel{\mathrm{\_}}{n}}}{(n-2)!}\right]=\frac{e(3-e)}{{(e-1)}^3}$$$$where{B}_{\stackrel{\mathrm{\_}}{n}}isthen-th{bernouli}^{\prime }snumber$$

Answered by witcher3 last updated on 09/Aug/23

$$\sum _{\mathrm{n}⩾0}{\mathrm{B}}_{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}=\frac{\mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}-1}=\mathrm{f}\left(\mathrm{x}\right)$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\sum _{\mathrm{n}⩾1}\frac{{\mathrm{x}}^{\mathrm{n}-1}}{(\mathrm{n}-1)!}{\mathrm{B}}_{\mathrm{n}}$$$${\mathrm{f}}^{″}\left(\mathrm{x}\right)=\sum _{\mathrm{n}⩾2}{\mathrm{B}}_{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}-2}}{(\mathrm{n}-2)!}$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{({\mathrm{e}}^{\mathrm{x}}-1)-{\mathrm{xe}}^{\mathrm{x}}}{{({\mathrm{e}}^{\mathrm{x}}-1)}^2}$$$${\mathrm{f}}^{″}\left(\mathrm{x}\right)=\frac{-{\mathrm{xe}}^{\mathrm{x}}{({\mathrm{e}}^{\mathrm{x}}-1)}^2-{2\mathrm{e}}^{\mathrm{x}}({\mathrm{e}}^{\mathrm{x}}-1)({\mathrm{e}}^{\mathrm{x}}-1-{\mathrm{xe}}^{\mathrm{x}})}{{({\mathrm{e}}^{\mathrm{x}}-1)}^4}$$$$=\frac{-{\mathrm{xe}}^{2\mathrm{x}}+{\mathrm{xe}}^{\mathrm{x}}-{2\mathrm{e}}^{2\mathrm{x}}+{2\mathrm{e}}^{\mathrm{x}}+{2\mathrm{xe}}^{2\mathrm{x}}}{{({\mathrm{e}}^{\mathrm{x}}-1)}^3}$$$$=\frac{{\mathrm{xe}}^{2\mathrm{x}}+{\mathrm{xe}}^{\mathrm{x}}-{2\mathrm{e}}^{2\mathrm{x}}+{2\mathrm{e}}^{\mathrm{x}}}{{({\mathrm{e}}^{\mathrm{x}}-1)}^3}$$$${\mathrm{f}}^{″}\left(1\right)=\frac{3\mathrm{e}-{\mathrm{e}}^2}{{(\mathrm{e}-1)}^3}=\sum _{\mathrm{n}⩾1}\frac{{\mathrm{B}}_{\mathrm{n}}}{(\mathrm{n}-2)!}$$$$$$

Commented by York12 last updated on 10/Aug/23

$$Icannotfindwords,butthankssomuch$$$$sir$$

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