$a_k = \sum\limits_{i=1}^{k-1}a_i \left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$ and $a_1 = 1$, prove the series is decreasing?

Question

If the series {$a_n$} is defined recursively in the following way:

1. $a_1 = 1$,
2. $a_k = \sum\limits_{i=1}^{k-1}a_i\cdot\left(\frac{1}{k-i}-\frac{1}{k-i+1}\right)$ for $k=2,\cdots, n$,

how can I prove that $a_k$ is decreasing?

Answer 1

Let $F(x) = \sum_{k=1}^\infty a_k x^k$ for $x \in (-1,1)$ (convergence is clear).

Claim: We have $F(x) = \frac{x}{(1-x^{-1})\log(1-x)}$ for $x \in (-1,1)$.

Proof: We have $F(x) = x+\sum_{k=2}^\infty \sum_{j=1}^{k-1} a_j \frac{1}{(k-j)(k-j+1)}x^k = x+\sum_{j=1}^\infty a_j x^j \sum_{k=j+1}^\infty \frac{x^{k-j}}{(k-j)(k-j+1)}$. Note the inner sum is $\sum_{r=1}^\infty \frac{x^r}{r(r+1)}$, so $F(x) = x+\left(\sum_{r=1}^\infty \frac{x^r}{r(r+1)}\right)F(x)$. The identity $1-\sum_{r=1}^\infty \frac{x^r}{r(r+1)} = (1-x^{-1})\log(1-x)$ thus gives the result. $\square$

Now, since $(1-x^{-1})F(x) = -1+\sum_{k=1}^\infty (a_k-a_{k+1})x^k$, we see $\frac{x}{\log(1-x)} = -1+\sum_{k=1}^\infty (a_k-a_{k+1})x^k$ for $x \in (-1,1)$. Changing $x \mapsto -x$, we have, for $x \in (-1,1)$, $$\frac{x}{\log(1+x)} = 1-\sum_{k=1}^\infty (-1)^k(a_k-a_{k+1})x^k.$$ We wish to prove $(-1)^k \frac{\partial^k}{\partial x^k}\left[\frac{x}{\log(1+x)}\right]\Big|_{x=0} \le 0$ for all $k \ge 0$. It then suffices to show $(-1)^k \frac{\partial^k}{\partial x^k}\left[\frac{x}{\log(1+x)}\right] \le 0$ on all of $[0,\infty)$, i.e., that $\frac{x}{\log(1+x)}$ is a Bernstein function. But this follows from the fact that $\log(1+x)$ is a complete Bernstein function together with the general result that $x/f(x)$ is a (complete) Bernstein function if $f(x)$ is.