Study the Convergence(pointwise and uniform ) of the sequence $(f_n)_{n \in \mathbb{N^*}}$
Where
$$f_n(z)=\sum_{n=1}^{n}\frac{1}{n^z}$$
Is the family $\{f_n : n \in \mathbb{N^*} \}$ normal ?
First we study the uniform convergence
We must show that $\sum_{n=1}^{\infty} \frac{1}{n^z}$ is uniformly convergence in $\text{Re}\ z \geq 1+ \epsilon$ where $\epsilon >0$
$$\frac{1}{n^z} = \frac{1}{e^{z\ \text {Log}\ n}} =\frac{1}{e^{x\ \text {Log}\ n}}\frac{1}{e^{i y\ \text {Log}\ n}} $$
Now $$|\frac{1}{n^z}| = \frac{1}{n^x} \leq \frac{1}{n^{\epsilon +1}} \ \ \text{if} \ \ x \geq 1 + \epsilon$$
Let $M_n = \sum_{n=1}^{\infty} \frac{1}{n^{\epsilon +1}}$ which is converge So $\sum_{n=1}^{\infty} \frac{1}{n^z} $ converges uniformly if $x \geq 1 + \epsilon$
Moreover since $$\sum_{n=1}^{\infty}|\frac{1}{n^z}|=\sum_{n=1}^{\infty}\frac{1}{n^x} \ \ \text{which converges } \ \ \forall \ x >1$$
So $\sum_{n=1}^{\infty} \frac{1}{n^z}$ is absolutely poinwise convergent for $x > 1$
Finally How to prove that the family is normal ?
Thank you .