Let
$f(x) = \begin{cases}\ln\frac{x}{e^x-1}, \quad x > 0; \\ 0, \quad\qquad x=0; \\ \ln\frac{x}{e^x-1}, \quad x < 0. \end{cases}$


Power series in 0:
$f(x) = \sum\_{n=1}^{\infty} a\_n x^n = -\frac{x}{2} - \frac{x^2}{24} + 
\frac{x^4}{2880} + \ldots$


I am interested in estimates (asymptotic) for the sum of the coefficients
$B(N) = \sum\_{n=1}^{N} a\_n$. For example, $|\ln \frac{1}{e-1} - B(N)| \ll \frac{1}{N}$.


Do you know of any references on this?


Thanks!



Estimate of the sum Taylor's coefficients
If $|z|\le 5/4$ then one may check that 
$$ 
\Big| \frac{e^{z}-1}{z} - 1 \Big| \le \sum\_{k=2}^{\infty} \frac{|z|^{k-1}}{k!} < 1,
$$ 
so that $\log (z/(e^z-1))$ (being the logarithm of a non-zero holomorphic function) is a holomorphic function in the region $|z|\le 5/4$ (indeed in a slightly bigger region). Therefore by the Cauchy estimates the coefficients $a\_n$ are bounded in size by $C (4/5)^n$ (integrate over a circle centered at $0$ of radius $5/4$) for a suitable constant $C$. Therefore 
$$ 
\Big| \log\frac{1}{e-1} - \sum\_{n\le N} a\_n \Big| = \Big| 
\sum\_{n=N+1}^{\infty} a\_n \Big| \le 5C (4/5)^{N+1}.
$$ 


