I have the following functional series $$ f(x)=\sum_{n\geq1}\frac{1}{n^x} $$ I need to find your domain, your continuity interval and where this function is differentiable.
My attemp:
Domain: the domain is $I:=(1,+\infty)$. Is easy to see that the series uniformly is convergent in $[\xi,+\infty)$ with $\xi>1$, because for each $\epsilon>0$ exists $N\in\mathbb{N}$ such that $$ \sum_{n\geq N}\frac{1}{n^\xi}<\epsilon $$ So, by comparing criteria, $f$ is uniformly convergent in $[\xi,+\infty)$ for all $\xi>1$. My first question is,Can I conclude thet $f$ is uniformly convergent in $I$?
Contuinity: As each $f_n(x)=\frac{1}{n^x}$ is continuous, for each $\xi>1$, $f$ is continuous in $[\xi,+\infty)$, so $f$ is continuous in $I$.
Differentiability: We have that $f_n'(x)=\frac{1}{n^x}ln(\frac{1}{n})=-\frac{1}{n^x}\ln(n)<-\frac{1}{n^{x-1}}$, then $\sum f_n'$ converges uniformly in any interval $[\xi,+\infty)$ for each $\xi>2$. (Again, with this can I conclude that $\sum f_n'$ is uniformly convergent in $(2,+\infty)$). So, $f$ is differentiable in $(2,+\infty)$.
Second question: My intervals are right?
This is not a geometric series; something like $\sum x^n$ would be. You are correct for the domain, because the integral test shows that $\sum n^{-x}$ converges if and only if $x > 1$. Furthermore, you can get uniform convergence on any $[\xi, \infty)$ for $\xi > 1$ by the Weierstrass $M$-test, but the convergence is not uniform on $[1, \infty)$ because $\lim_{x \to 1+} \sum n^{-x} = \infty$.
As far as differentiability, you can get it on $(1, \infty)$ by being more judicious with your estimate of the logarithm. In particular, for any $\epsilon > 0$ we have
$$\ln(n) \le n^{\epsilon}$$
once $n$ is sufficiently large (in terms of $\epsilon$). Furthermore, the derivatives uniformly converge on $[\xi, \infty)$ for any $\xi > 1$, again by the $M$-test.
In fact, a lot more is true. Not only is $\sum n^{-x}$ infinitely differentiable on $(1, \infty)$, each of the series approximating the derivatives are uniformly convergent on any interval $[1 + \epsilon, \infty)$.