# Supremum of the function of a sum for Weierstrass M-test

I have to prove the uniform convergence of $\sum_{k=1}^\infty \frac{k+z}{k^3 + 1}$ on the closed disc $D_1(0)$. Using the M-test, $|\frac{k+z}{k^3 +1}| \leq |\frac{k+1}{k^3 +1}| = \frac{1}{k^2 - k + 1}$. I'm a little stuck here with what to pick as my $M_k$ value. I'd assume that $\sum \frac{1}{k^2 - k +1}$ converges, but I don't know what to compare it to. Tips would be appreciated.

One approach would be to rearrange $\sum_{k=1}^\infty\frac{1}{k^2-k+1}$as $$1+\sum_{k=1}^\infty\frac{1}{(k+1)^2-(k+1)+1}=1+\sum_{k=1}^\infty\frac{1}{k^2+k+1}$$ and now you can compare $$\sum_{k=1}^\infty\frac{1}{k^2+k+1}\leq\sum_{k=1}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}$$ so the original series is bounded above by $1+\frac{\pi^2}{6}$.