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

p r o v e t h a t

\underset{n = 2}{\sum^{\infty}} [\frac{B_{\overline{n}}}{(n - 2)!}] = \frac{e (3 - e)}{{(e - 1)}^{3}}

w h e r e B_{\overline{n}} i s t h e n - t h {b e r n o u l i}^{'} s n u m b e r

Answered by witcher3 last updated on 09/Aug/23

\sum_{n ⩾ 0} B_{n} \frac{x^{n}}{n!} = \frac{x}{e^{x} - 1} = f (x)

f^{'} (x) = \sum_{n ⩾ 1} \frac{x^{n - 1}}{(n - 1)!} B_{n}

f^{″} (x) = \sum_{n ⩾ 2} B_{n} \frac{x^{n - 2}}{(n - 2)!}

f^{'} (x) = \frac{(e^{x} - 1) - {xe}^{x}}{{(e^{x} - 1)}^{2}}

f^{″} (x) = \frac{- {xe}^{x} {(e^{x} - 1)}^{2} - {2 e}^{x} (e^{x} - 1) (e^{x} - 1 - {xe}^{x})}{{(e^{x} - 1)}^{4}}

= \frac{- {xe}^{2 x} + {xe}^{x} - {2 e}^{2 x} + {2 e}^{x} + {2 xe}^{2 x}}{{(e^{x} - 1)}^{3}}

= \frac{{xe}^{2 x} + {xe}^{x} - {2 e}^{2 x} + {2 e}^{x}}{{(e^{x} - 1)}^{3}}

f^{″} (1) = \frac{3 e - e^{2}}{{(e - 1)}^{3}} = \sum_{n ⩾ 1} \frac{B_{n}}{(n - 2)!}

Commented by York12 last updated on 10/Aug/23

I c a n n o t f i n d w o r d s, b u t t h a n k s s o m u c h

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