1. We know that in quantum field theory we include infinities at each order of the perturbative expansion of the renormalization $Z$ factors about the coupling constant $\lambda$ to absorb the divergences of the loop diagrams, so does this mean that $Z$ must be infinite as well?

2. On the other hand, if we turn the coupling constant $\lambda$ to zero, the interacting theory then becomes a free theory, so the $Z$ must be $1$ in this case. This means that the $Z$ should be a small variation of $1$ when $\lambda$ is small.

Combining the above two arguments, does it mean that although there are infinities in each $\lambda^n$ order term in the expansion of $Z$, their total sum turns out to be a finite number which is a small variation around $1$?

So, when people say the renormalization $Z$ factors are infinite, do they actually mean that the $Z$s are infinite at each order?

1. We know that in quantum field theory we include infinities at each order of the perturbative expansion of the renormalization $Z$ factors about the coupling constant $\lambda$ to absorb the divergences of the loop diagrams, so it seems $Z$ must be infinite.

2. On the other hand, if we turn the coupling constant $\lambda$ to zero, the interacting theory then becomes a free theory, so the $Z$ must be $1$ in this case. This means that the $Z$ should be a small variation of $1$ when $\lambda$ is small.

3. Moreover, according to the Kallen-Lehmann spectral form we must have $Z \in [0, 1]$.

Combining the above arguments, does it mean that although there are infinities in each $\lambda^n$ order term in the expansion of $Z$, their total sum turns out to be a finite number which is a small variation around $1$? That is, when people say the renormalization $Z$ factors are infinite, do they actually mean that the $Z$'s are infinite at each order?

1.We know that in quantum field theory we include infinities at each order of the perturbative expansion of the renormalization $Z$ factors about the coupling constant $\lambda$ to absorb the divergences of the loop diagrams, so does this mean that $Z$ must be infinite as well?

2.On the other hand, if we turn the coupling constant $\lambda$ to zero, the interacting theory then becomes a free theory, so the $Z$ must be $1$ in this case. This means that the $Z$ should be a small variation of $1$ when $\lambda$ is small.

Combining the above two arguments, does it mean that although there are infinities in each $\lambda^n$ order term in the expansion of $Z$, their total sum turns out to be a finite number which is a small variation around $1$?

So, when people say the renormalization $Z$ factors are infinite, do they actually mean that the $Z$s are infinite at each order?

1. We know that in quantum field theory we include infinities at each order of the perturbative expansion of the renormalization $Z$ factors about the coupling constant $\lambda$ to absorb the divergences of the loop diagrams, so does this mean that $Z$ must be infinite as well?

2. On the other hand, if we turn the coupling constant $\lambda$ to zero, the interacting theory then becomes a free theory, so the $Z$ must be $1$ in this case. This means that the $Z$ should be a small variation of $1$ when $\lambda$ is small.

Combining the above two arguments, does it mean that although there are infinities in each $\lambda^n$ order term in the expansion of $Z$, their total sum turns out to be a finite number which is a small variation around $1$?

So, when people say the renormalization $Z$ factors are infinite, do they actually mean that the $Z$s are infinite at each order?