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Let: $f(z)=\frac{z}{e^z-1}$ if $z\ne0$, and $f(z)=1$ if $z=0$.

Please help to prove that $\sum_{k=0}^{n-1}\binom n kf^{(k)}(0)=0$ for any $n>1$ and $f^{(2n+1)}(0)=0$ for any natrual $n$.

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up vote 1 down vote accepted

HINT about the sum.

  1. Start with $f(z) \left(\exp(z)-1\right) = z$.

  2. Write Taylor series for $f(z)$ and for $\mathrm{e}^z-1$ and multiply them.

  3. The coefficients of the resulting series will be Cauchy sums of series coefficients of $f(z)$ and $\mathrm{e}^z-1$.

  4. Now compare to the right-hand-side $z$.

The other property follows from: $$ f(-z) = \frac{-z}{\exp(-z)-1} = -\frac{z \exp(z)}{1-\exp(z)} = z + \frac{z}{\exp(z)-1} = z + f(z) $$ now take the derivative of order greater than one for the both sides at $z=0$.

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