= Solution
Fix $k\geq1$ and write $n=qk+r$ with $1\leq r\leq k$ and $q\geq0$. Repeatedly use the given inequality with its second index equal to $k$. This yields
$$
x_{qk+r}\leq x_r+q(x_k+\alpha_k).
$$
There are only finitely many possible remainders for fixed $k$, so
$$
\limsup_{n\to\infty}\frac{x_n}{n}\leq\frac{x_k+\alpha_k}{k}.
$$
Taking $k=1$ rules out positive infinity for this upper limit. Let $L=\liminf x_n/n$, which may be negative infinity, and choose $k_j\to\infty$ with $x_{k_j}/k_j\to L$. The assumption $\alpha_{k_j}/k_j\to0$ then gives $\limsup x_n/n\leq L$. This also works when $L=-\infty$, by taking an arbitrarily negative upper bound. Therefore the <asymmetrically almost-subadditive sequence> has
$$
\boxed{\lim_{n\to\infty}\frac{x_n}{n}=\gamma\in[-\infty,\infty),\qquad\gamma=\inf_{k\geq1}\frac{x_k+\alpha_k}{k}}.
$$
The last infimum identity follows from the fixed-$k$ bound and from the convergence of the corrected ratios to $\gamma$. No positivity assumption on $\alpha_n$ is required.
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