Show that $\sum_{k=0}^\infty \frac{1}{k^2-z}$ converges uniformly by Weierstrass M Test 
Show that for each $r>0$ the series $$\sum_{k=0}^\infty \frac{1}{k^2-z}$$ converges uniformly on the set $E_r=\{z\mid |z|\leq r, z\neq k^2 \,\,\text{for} \,\,k=0,1,2,\cdots\}$

I wish to use the Weierstrass M-Test to show that this series converges uniformly. My intuition is to use $M_k = \frac{1}{k^2-r}$, but this is not necessarily positive, so I'm stuck.
Any hints as to picking an $M_k$ would be most appreciated. Then I will continue to show that all the conditions of the M-Test are satisfied.
UPDATE: The question does not specifically state to use the Weierstrass M-Test, but it was how the preceding few questions were solved. Any other methods of showing uniform convergence are welcome.
 A: With the help of Daniel Fischer identifying my mistake, I use the Weierstrass M-Test with $M_k = \frac{1}{k^2-r}$. 
First, we assume $k^2>r$, as this doesn't change the test for uniform convergence, so that $M_k$ are positive terms. 
Then, we need to ensure that $|f_k(z)|\leq M_k$ for $z\in E_r$. We have $|k^2-z|\geq |k^2|-|z| \geq k^2-r$ by the Triangle Inequality and definition of $E_r$. Thus, $$\frac{1}{k^2-z}\leq \frac{1}{k^2-r} = M_k$$ so the condition is satisfied.
Lastly, we need $$\sum_{k=0}^\infty M_k = \sum_{k=0}^\infty \frac{1}{k^2-r}$$ to converge for all $r>0$. To show this, we use the limit comparison test with $b_k = \sum_{k=1}^\infty M_k$ and $a_k = \sum_{k=1}^\infty \frac{1}{k^2}$. Then,
$$c=\lim_{k \to \infty} \left(\frac{1}{k^2}\right)\left(\frac{k^2-r}{1}\right) = 1.$$ Since $0<c<\infty$, both series either converge or diverge. We know $\sum a_k$ converges since it is a $p$-series with $p=2$. Thus, $\sum M_k$ converges.
We can therefore apply the Weierstrass M-Test to conclude that $\sum_{k=0}^\infty \frac{1}{k^2-z}$ converges uniformly on $E_r$.
