Comparing first, second, and third ionization energies between atoms
If you know the trends for $1^\text{st}$ ionization energies, a good first approximation for the trends of $n^\text{th}$ ionization energies at any given point is the $1^\text{st}$ ionization trend $n$ elements earlier. 

As an example, look at the first and second ionization energies for the second row elements (Table).
From $\ce{Li}$ to $\ce{Be}$, the $1^\text{st}$ ionization energy increases due to the greater effective charge on the outermost electron. From $\ce{Be}$ to $\ce{B}$, the $1^\text{st}$ ionization energy slightly decreases, as the electron we removing is in the p-subshell and is shielded by the inner s-subshell electrons. From $\ce{B}$ to $\ce{C}$, it increases again.

What about the $2^\text{nd}$ ionizations? Well from $\ce{Be}$ to $\ce{B}$, it increase, as the $2^\text{nd}$ electron from each is pulled from the s-subshell, but $\ce{B^+}$ has a greater effective charge than $\ce{Be^+}$. From $\ce{B}$ to $\ce{C}$, it decreases, as the $2^\text{nd}$ electron pulled from $\ce{C}$ is still in the less shielded p-subshell. 

The $\ce{Be}$ to $\ce{B}$ trend in $2^\text{nd}$ ionization matches up with the $\ce{Li}$ to $\ce{Be}$ trend in $1^\text{st}$ ionization. The same is true for $\ce{B}$ to $\ce{C}$ (2nd) and $\ce{Be}$ to $\ce{B}$ (1st).