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The second equality in (6.88) he says was obtained by expanding the denomitator by the binomial theorem. It is probably very dumb but I'm not following. I see how the 1 and the vacuum term in the numerator cancel with the denominator and give a 1. But I don't follow how he got the rest.

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The expansion is $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$ and you only get the linear term to order $g^1$. –  Michael Brown Feb 21 '13 at 12:53
    
I see, that answers my question. Thanks. You should have posted it as an answer tho. –  Barefeg Feb 21 '13 at 13:07
    
Done, with a little expansion. –  Michael Brown Feb 21 '13 at 13:12

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The expansion is

$$ \frac{1}{1-x} = 1 + x + x^2 + \cdots $$

where $x$ is the vacuum diagram. You only get the linear term to first order in $g$, which cancels the vacuum diagram in the numerator. There is a combinatorial proof that the cancellation of vacuum diagrams holds to all orders - Ryder should have it.

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Just completing with a reference, in case it's not in the Ryder, Peskin & Schroeder has the demonstration in the fourth chapter (the one developing the perturbation tools). –  Learning is a mess Feb 21 '13 at 13:36

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