Cancelling terms in an infinite series over $\mathbb{N}^2$

Question

I try to understand Don Zagiers seemingly simple proof of the recurrence relation \begin{align*} \sum_{0<j<k,\ j\ \text{even}}\zeta(j)\zeta(k-j)=\frac{k+1}{2}\zeta(k)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k\geq4\text{ even})\ . \end{align*} I figured out everything but one step: \begin{align} \sum_{m,n>0}f(m,n)-f(m+n,n)-f(m,m+n)=\sum_{n>0}f(n,n) \end{align} where (for the case k=4) \begin{align*} f(m,n)=\frac{1}{mn^3}+\frac{1}{2m^2n^2}+\frac{1}{m^3n}\ . \end{align*} If $\ \sum_{m,n>0}f(m,n)\ $ existed, you could split up the sum and you would easily see that the sum over $f(m,n)$ with $m\neq n$ cancels with the sums over $f(m+n,n)$ and $f(m,m+n)$. Zagier illustrates this argument by \begin{align*} \sum_{m,n>0}f(m,n)-f(m+n,n)-f(m,m+n)&=\left(\sum_{m,n>0}-\sum_{m>n>0}-\sum_{n>m>0}\right)f(m,n)\\&=\sum_{n>0}f(n,n) \end{align*} However the sums over the single terms do not exist - due to the divergence of the harmonic series. So what am I missing here?