Through the photoelectric effect among many others we learn that light is actually comprised of discreet quanta of energy. That's because of the energy of the emitted electrons as well as the minimum frequency of incident light that is able to start the process. A photon of frequency (energy) less than the threshold frequency is not able to free the electron from the surface. However if the required energy to free the electron was greater than this frequency by an integral value why couldn't the electron simply absorb two photons or three photons (or any required number of photons) so as to be released? In this case even with frequencies lesser than the threshold frequency there would be photoelectric effect.
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This multi-photon process are possible, but of higher order (in the couplage $\alpha$ between the photons and the electrons), and therefore highly improbable. Furthermore, you need that the frequency of the photons $\nu$ is such that $n \nu=W$, with $W$ the energy needed to eject an electron from the metal, et $n$ an integer. But in principle, you are right, there should be a (very small) photoelectric current scaling as $\alpha^n$ when $n\nu= W$. I don't know if people have tried to measure it.
EDIT: 5 minutes of googling shows that there are plenty of references about the multi-photons photoelectric effect. Here is a random selection: one experiment (free access), another.
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$\begingroup$ Ohh thanks! I had searched on Google but I was finding things relating to lasers that I did not quite get the relationship of. Like stimulated emission in lasers. I did not understand the couplage of photons and electrons of your answer! $\endgroup$ Commented Jun 12, 2014 at 3:07
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$\begingroup$ @user43470: Nothing fancy here: you need an interaction between the electron and the photons for the photons to be absorbed. This coupling is proportional to the electric charge. $\endgroup$– AdamCommented Jun 12, 2014 at 13:33