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## Homework Statement

I have the Lagrangian $$L=-\frac{1}{2}\phi\Box \phi-\frac{1}{2}m^2\phi^2$$ and I need to show that the propagator in the momentum space I obtain using this lagrangian (considering no interaction) is the same as if I consider the free Lagrangian to be $$L_{free}=-\frac{1}{2}\phi\Box \phi$$ and treat the mass term as an interaction $$L_{int}= -\frac{1}{2}m^2\phi^2$$

## Homework Equations

## The Attempt at a Solution

So in the normal case the propagator for a mass m scalar particle is $$\frac{i}{p^2-m^2+i\epsilon}$$. For the other approach I get that the propagator looks like this: $$\frac{1}{p^2+i\epsilon}-im^2(\frac{1}{p^2+i\epsilon})^2-m^4(\frac{1}{p^2+i\epsilon})^3+im^6(\frac{1}{p^2+i\epsilon})^4+...$$ But i am not sure how to show they are equal. Can someone help me?