The question is simple: for integer $k\geq 2$, how to calculate the power series expansion of $f(x)=\frac{kx}{(x+1)^k-1}$ at $x=0$?
As far as I am aware, there are two effective (operatable) ways to do that:
- Using Taylor expansion: just calculate the derivative of any order at $x=0$. The derivative becomes hard to calculate when the order gets larger and larger.
- One can expand $f(x)$ as follows: $$f(x)=\frac{kx}{\sum_{i=1}^k\binom{k}{i}x^i}=\frac{1}{1+\sum_{i=2}^k\frac{\binom{k}{i}}{k}x^{i-1}}=\sum_{j=0}^\infty \left(-x\sum_{i=0}^{k-2}\frac{\binom{k}{i+2}}{k}x^i\right)^j.$$ This expansion is well-defined since the degree of $\left(-x\sum_{i=0}^{k-2}\frac{\binom{k}{i+2}}{k}x^i\right)^j$ tends to infinity when $j\rightarrow \infty$. Then the coefficient of $x^N$ in the expansion of $f(x)$ follow from calculating the coefficient of $x^N$ in the expansion of $\sum_{j=0}^N \left(-x\sum_{i=0}^{k-2}\frac{\binom{k}{i+2}}{k}x^i\right)^j$, which is a finite sum. However, it is still hard to do the real calculation, since the result will involve tons of summation involving binomial and multinomial coefficients.
So, is there any closed-form, or, at least a simpler formula for the coefficients in the power series expansion of $f(x)$?
The power series about $x=0$ is given by $$ \frac{{kx}}{{(x + 1)^k - 1}} = \sum\limits_{n = 0}^\infty {k^n \beta _n (1/k)\frac{{x^n }}{{n!}}} , $$ where the $\beta_n(\lambda)$ are the degenerate Bernoulli numbers.