How does the isothermal expansion of a gas increase entropy of surroundings? I was under the impression that if a gas expands isothermally then energy must be transferred from the surroundings to the gas as it is performing work. But in that case surely using $$\Delta S= \frac{\Delta Q}{T}$$ if the change in energy for the surroundings is negative then the change in entropy must also be negative?
 A: The $Q$ term that you used in your formula, represent the heat absorbed (or evolved) for reversible processes only. For irreversible processes the term for change in entropy is different.
In an isothermal process,
$\Delta T = 0 \Rightarrow \Delta U =0$,
Therefore, $P\Delta V = q$
When the gas expands against external pressure it uses some of its internal energy and to compensate for the loss in the internal energy it absorbs heat from the surrounding.
But the thing about reversible processes is that, $\Delta {S_{universe}}=0$
$\Delta {S_{system}}=-\Delta {S_{surrounding}}$.
For all irreversible processes, the entropy of the universe increases. It doesn't matter if the surrounding's entropy decreases and if it does, the entropy change for the universe will either be 0 (reversible processes) or positive (irreversible processes) .
For irreversible processes, the entropy change associated with the state change is
$dS=\frac{Q_{actual}}{T}+\frac{(dW_{reversible}-dW_{actual})}{T}$
The subscript 'actual' refers to an actual process i.e, irreversible process.
Since, $dW_{reversible} > dW_{actual}$
$dS > \frac{dQ_{actual}}{T}$.
For more check this out : http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node48.html
