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There is a explicit form that admit Bernoulli numbers rationals but there is another definition where the Bernoulli numbers are $B_n$, such that $\displaystyle \frac{x}{e^x-1}= \sum_{n=0}^\infty B_n \frac{x^n}{n!}$.

How can I prove that?? or it's equivalence?

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  • $\begingroup$ You should tell us what that "explicit form" is, as without knowing that one cannot show it is equivalent to the other definition. $\endgroup$ – Torsten Schoeneberg Sep 4 '19 at 3:25
  • $\begingroup$ For a quick explanation, the derivatives on the left continued to 0 have rational values at 0 since there are no irrational terms showing up aside from $e^x$, which either becomes 1 or gets used up in $\lim_{x\to0}\frac{e^x-1}x=1$, both of which are rational. $\endgroup$ – Simply Beautiful Art Sep 4 '19 at 3:27
  • $\begingroup$ @Torsten I presume they mean to say some forms admit the solution trivially, though they want to prove it directly from the generating function definition. $\endgroup$ – Simply Beautiful Art Sep 4 '19 at 3:29
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    $\begingroup$ If you write $x=\sum\limits_{n=0}^\infty B_n \dfrac {x^n}{n!} (e^x-1)$ and expand $(e^x-1)$ and equate coefficients of like powers of $x$, won't you get a recurrence relation for the Bernoulli numbers, where the coefficients are rational? $\endgroup$ – J. W. Tanner Sep 4 '19 at 3:32
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We have $$\frac{x}{e^x-1} = \frac{x}{x + x^2/2 + x^3/6 + x^4/24 + \cdots} = \frac{1}{1 + x/2 + x^2/6 + x^3/24 + \cdots}.$$ Now, if we let $f(x) := \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots$, then $$\frac{x}{e^x-1} = \frac{1}{1 + f(x)} = 1 - f(x) + (f(x))^2 - (f(x))^3 - \cdots.$$ In this expansion, each power of $f(x)$ has rational coefficients, and since $f(x)$ is divisible by $x$, the final coefficient of $x^n$ is the same as the coefficient of $x^n$ in $1 - f(x) + (f(x))^2 - (f(x))^3 + \cdots + (-1)^n (f(x))^n$.

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From the definition you gave for Bernoulli numbers, $$x=\left(\sum\limits_{n=0}^{\infty} B_n\dfrac{x^n}{n!}\right)(e^x-1)=\left(\sum\limits_{n=0}^{\infty} B_n\dfrac{x^n}{n!}\right)\left(\left(\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\right)-1\right)$$

$$=\sum\limits_{n=0}^{\infty}\left(\left(\sum\limits_{j=0}^n\dfrac{B_j}{j!(n-j)!}\right)-\dfrac{B_n}{n!}\right)x^n.$$

Equating the coefficients of $x$ shows that $B_0=1,$

and equating the coefficients of $x^{n+1}$ for $n>0$ shows that $$0=\left(\sum\limits_{j=0}^{n+1}\dfrac{B_j}{j!(n+1-j)!}\right)-\dfrac{B_{n+1}}{(n+1)!}$$

i.e., $$B_{n+1}=\sum\limits_{j=0}^{n+1} \binom{n+1}j{B_j}.$$

This recurrence relation can be used to calculate $B_n$ for $n>0$,

since $B_{n+1}$ can be cancelled from both sides.

In any event, it is clear from this that the Bernoulli numbers are rational.

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