 Accumulating asymptotic exponents obstruct Mellin continuation

Strict increase alone does not suffice in the preceding theorem. Let ${\sigma }_j=1-1/j$, $c_j=2^{-j}$ and $f(y)=\eta (y){\sum }_{j\ge 1}{2}^{-j}{y}^{1-1/j}$, where $\eta$ is continuous, equals one on $(0,1]$ and vanishes on $[2,\infty )$. After division by the next power, every finite remainder extends continuously to zero. Nevertheless the Mellin transform has genuine poles at $-{\sigma }_j$ accumulating at $-1$, so it cannot be a meromorphic function on the whole plane.