Proving an identity for Bernoulli polynomials

sequences-and-series calculus bernoulli-numbers

Solution 1:

Hint. You may just write $$ \frac{te^{(1-x)t}}{e^t-1}=\frac{t\:e^{t}e^{x(-t)}}{e^t(1-e^{-t})}=\frac{t\:e^{x(-t)}}{1-e^{-t}}=\frac{(-t)e^{x(-t)}}{e^{(-t)}-1} $$ and use the definition.

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