For every nonzero integer $m$ and every integer $n \ge 1$, the $n$\textsuperscript{th} harmonic number $H_n = 1 + \tfrac12 + \dots + \tfrac1n$ satisfies the identity \[ H_n \;=\; \frac{1}{m}\,\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}\, \binom{m k}{k}\binom{n + (m-1)k}{n - k}. \] The cases $m = 1$ and $m = 2$ are classical; for general nonzero integer $m$ the identity was conjectured by P.~Bala in the OEIS entry A001008 in 2022 and remained open. We prove it here, working throughout in $\QQ[[x]]$. The proof reduces, via a substitution $u = x/(1-x)^m$, to two formal-power-series identities: a Lagrange--Bürmann evaluation of $\sum_{k\ge1} \binom{mk}{k} u^k / k$, and the fixed-point fact that under that substitution the unique solution $v(u)$ of $v = u(1-v)^{m}$ is $v = x$. The argument extends verbatim to arbitrary complex $m \ne 0$.

翻译：对每个非零整数 $m$ 及每个整数 $n \ge 1$，第 $n$ 个调和数 $H_n = 1 + \tfrac12 + \dots + \tfrac1n$ 满足恒等式 \[ H_n \;=\; \frac{1}{m}\,\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}\, \binom{m k}{k}\binom{n + (m-1)k}{n - k}. \] 情形 $m = 1$ 与 $m = 2$ 是经典的；对于一般非零整数 $m$，该恒等式由 P.~Bala 于 2022 年在 OEIS 条目 A001008 中提出猜想，并一直悬而未决。我们在此证明该猜想，全程在 $\QQ[[x]]$ 中展开工作。该证明通过代换 $u = x/(1-x)^m$ 简化为两个形式幂级数恒等式：关于 $\sum_{k\ge1} \binom{mk}{k} u^k / k$ 的 Lagrange--Bürmann 求值，以及在该代换下方程 $v = u(1-v)^{m}$ 的唯一解 $v(u)$ 满足 $v = x$ 这一不动点事实。该论证可直接推广至任意非零复数 $m$。