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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!

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    $\begingroup$ Is $f(x)$ correctly defined? If $x<0$, then $\frac {-x} {e^x-1} < 0$. $\endgroup$
    – user64494
    Commented Oct 25, 2014 at 15:23

1 Answer 1

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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 $$ \Big| \log\frac{1}{e-1} - \sum_{n\le N} a_n \Big| = \Big| \sum_{n=N+1}^{\infty} a_n \Big| \le 5C (4/5)^{N+1}. $$

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