Euler series transformation

Question

Consider an infinite power series \begin{equation} S(x) = \sum\limits_{n=0}^\infty b_n x^n \end{equation} If the coefficients of that series can be written as an $n$th backwards difference of some other set of coefficients then the series can be resummed. In other words the following identity holds: \begin{equation} S(x) = \frac{1}{1+x} \sum\limits_{k=0}^\infty a_k \left(\frac{x}{1+x}\right)^k \end{equation} where $\Delta^n a_n := \sum\limits_{k=0}^n \binom{n}{k} (-1)^{n-k} a_k = b_n$ or $a_n = \sum\limits_{k=0}^n \binom{n}{k} b_k$.

Now, let us assume that \begin{equation} a_n = \frac{H^{(m_1)}_{n+1}}{(n+1)^m} \end{equation} where $m$ and $m_1$ are non-negative integers and $H^{(m_1)}_{n+1}$ are generalized Harmonic numbers.

How do I compute the coefficients $\left\{b_n\right\}_{n=0}^\infty$ in this case?

Hint: In Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$ the answer is found for the case $(m,m_1)=(1,2)$.

Answer 1

Let us consider the case $m_1=0$ at first. In other words we take: \begin{equation} a^{(m)}_n = \frac{1}{(n+1)^m} \end{equation} where $m \ge 1$ is a fixed integer. By generalizing the methods from the question cited above (or by using the trick $1/(k+1)^m = \int\limits_0^\infty \theta^{m-1}/(m-1)! \exp[-\theta (k+1)] d\theta$ we easily get: \begin{equation} b^{(m)}_n = \frac{(-1)^n}{n+1} \frac{S_{m-1}(n)}{(m-1)!} \end{equation} where \begin{equation} S_m(n) = \left\{ \begin{array}{rr} H_{n+1} & \mbox{if $m=1$}\\ H_{n+1}^2 + H^{(2)}_{n+1} & \mbox{if $m=2$}\\ H_{n+1}^3 + 3 H_{n+1} H_{n+1}^{(2)}+ 2 H^{(3)}_{n+1} & \mbox{if $m=3$}\\ H_{n+1}^4 + 6 H_{n+1}^2 H_{n+1}^{(2)} +3 (H_{n+1}^{(2)})^2 +8 H_{n+1} H_{n+1}^{(3)}+ 6 H^{(4)}_{n+1} & \mbox{if $m=4$}\\ \vdots \end{array} \right. \end{equation} and in general: \begin{equation} S_m(n) = \sum\limits_{\lambda=1}^m \frac{1}{\lambda!} \sum\limits_{\begin{array}{r} p_1+\cdots+p_\lambda=m \\ p_1 \ge 1, \cdots,p_\lambda\ge 1\end{array}} \prod\limits_{q=1}^\lambda \frac{H_{n+1}^{(p_q)}}{p_q} \end{equation} The Euler series transformation results in this case in a sequence of following identities: \begin{equation} \frac{1}{x} Li_m\left( \frac{x}{x+1} \right) = \sum\limits_{n=0}^\infty b^{(m)}_n x^n \end{equation} In particular: \begin{eqnarray} Li_1\left( \frac{x}{x+1}\right) &=& - \frac{1}{0!} Li_1(-x) \quad \\ Li_2\left( \frac{x}{x+1}\right) &=& - \frac{1}{1!} \sum\limits_{n=1}^\infty \frac{H_n}{n} (-x)^n \\ Li_3\left( \frac{x}{x+1}\right) &=& - \frac{1}{2!}\sum\limits_{n=1}^\infty \frac{H_n^2+H_n^{(2)}}{n} (-x)^n \\ Li_4\left( \frac{x}{x+1}\right) &=& - \frac{1}{3!}\sum\limits_{n=1}^\infty \frac{H_n^3+3 H_n H_n^{(2)}+2 H_n^{(3)}}{n} (-x)^n \\ \vdots \end{eqnarray} Now we consider the case $m=0$. In other words we take: \begin{equation} a^{(m_1)}_n = H^{(m_1)}_{n+1} \end{equation} where $m_1 \ge 1$ is a fixed integer. Using the same methods as above we establish the following: \begin{equation} b^{(m_1)}_n = \frac{(-1)^{n-1}}{n(n+1)} \frac{\bar{S}_{m-1}(n)}{(m-1)!} \end{equation} where \begin{equation} \bar{S}_m(n) = S_m(n) - S_{m-1}(n) \end{equation} This leads to following identities: \begin{equation} \frac{x+1}{x} Li_m\left( \frac{x}{x+1} \right) = \sum\limits_{n=0}^\infty b^{(m_1)}_n x^n \end{equation} In particular: \begin{eqnarray} Li_1\left( \frac{x}{x+1} \right) &=& \log(1+x) \\ Li_2\left(\frac{x}{x+1}\right) &=& -Li_2\left(\frac{1}{x+1}\right)-\log ^2(x+1)+\log (x) \log (x+1)+\frac{\pi ^2}{6} \\ Li_3\left(\frac{x}{x+1}\right) &=& \frac{x}{x+1}\left(1+\frac{1}{2}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n (n+1)} \left(H_{n+1}^2+H_{n+1}^{(2)}-2 H_{n+1}\right) x^n \right) \\ \vdots \end{eqnarray}