<s>[INST] Is $f_n(z) = \dfrac {2nz - 1} {z + n^2}$ uniformly convergent on the disc $|z| <1$? The pointwise limit of $f_n(z) = \dfrac {2nz - 1} {z + n^2}$ is $0$.
By the triangle inequality,
$\Bigg|\dfrac {2nz - 1} {z + n^2}\Bigg| \le \Bigg|\dfrac {2nz} {z + n^2}\Bigg| + \Bigg|\dfrac {1} {z + n^2}\Bigg|$
On the disc $|z| <1$, $\Bigg|\dfrac {1} {z + n^2}\Bigg|\le \dfrac {1} {n^2} \le \dfrac {1} {n}$. However I cannot seem to find bounds for $\Bigg|\dfrac {2nz} {z + n^2}\Bigg|$.
I'm beginning to think that the sequence is only pointwise convergent in the specified disc.
 [/INST] No, $|1/(z+n^2)| \le 1/n^2$ fails. Try $z=-1/2$ for example. Instead note that for $n>1,$
$$\left |\frac{2nz-1}{z+n^2}\right | \le \frac{|2nz|+1}{n^2 - |z|} \le \frac{2n+1}{n^2 - 1}.$$
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