= Solution
For a fixed number $N$ of identical bulbs, first approximate each as a <resistor> of <electrical resistance> $R$ connected to a fixed ideal battery <electric potential difference> $V$. A <series circuit> has <electrical resistance> $NR$, while a <parallel circuit> has <electrical resistance> $R/N$. Thus
$$
P_{\rm series}=\frac{V^2}{NR},\qquad P_{\rm parallel}=\frac{NV^2}{R},\qquad\boxed{\frac{P_{\rm parallel}}{P_{\rm series}}=N^2}.
$$
Parallel connection therefore maximizes dissipation in this model: every bulb receives the full source <electric potential difference> instead of sharing it. For $N=1$ the two arrangements coincide.
An actual <incandescent light bulb> changes <electrical resistance> as its filament heats, so a single constant $R$ need not describe both operating points. The same qualitative conclusion holds for identical stable bulbs with increasing current-voltage function $i(U)$: parallel operation gives $P_p=NV i(V)$, whereas symmetric series operation gives $P_s=V i(V/N)$. The former is larger for $N>1$. This comparison assumes a fixed collection and an ideal source as specified; an unlimited number of parallel bulbs on a zero-internal-resistance ideal battery would have no finite total-power maximum.
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