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We derive a first-order recurrence for \(a_n(t) = \sum_{k=0}^{n} \frac{(-1)^{n-k}}{1+tk} \binom{n}{k}\) (\(t\) fixed \(t\neq -\frac{1}{m}, m\in \mathbb{N}\)). The first-order recurrence yields an alternative proof for Riordan’s theorem: \(a_n(t) = \binom{1/{t+n}}{n}^{-1}(-1)^n\) and also yields the ordinary generating function \(\sum_{n=0}^{\infty} a_n(t) x^n\) for \(t \in \mathbb{N}.\)From the latter, one easily computes \(\sum_{n=0}^{\infty}a_n(t)\) which turns out to be the well-known \(\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} = \ln 2\) for \(t=1\). For \(t=2\), we get \(\sum_{n=0}^{\infty} (-1)^n\frac{2n}{(2n+1)} = \frac{\ln(\sqrt{2}+1)}{\sqrt{2}}\).