I am confused with how to get from $$\frac{x}{e^{x}-1}=\sum_{n=0}^{\infty}B_{n}\frac{x^{n}}{n!}$$
In the book im studying from it says its a definition but idk how to generate terms from the right hand side or anything. I guess i should expand the left side in a series but idk what to do really. Thank you for your help
Indeed, the definition of $(B_n)_{n\geqslant 0}$ is the relation $\displaystyle\frac{x}{e^x-1}=\sum_{n=0}^{+\infty}\frac{B_n}{n!}x^n$. Therefore $$ x=\frac{x}{e^x-1}(e^x-1)=\left(\sum_{n=0}^{+\infty}\frac{B_n}{n!}x^n\right)\left(\sum_{n=1}^{+\infty}\frac{x^n}{n!}\right)=\sum_{n=1}^{+\infty}\left(\sum_{k=0}^{n-1} \frac{B_k}{k!(n-k)!}\right)x^n $$ Thus $B_0=1$ and for all $n\geqslant 1$, $$ B_n=-n!\sum_{k=0}^{n-1}\frac{B_k}{k!(n-k+1)!}=-\frac{1}{n+1}\sum_{k=0}^{n-1}\binom{n+1}{k}B_k $$ This recursive relation makes it possible to compute the Bernoulli numbers.