Consider the following theorem:
If $\sum_{n=0}^\infty a_n x^n $ converges for all $x \in (-R,R)$ then the differentiated series $\sum_{n=0}^\infty n a_n x^{n-1}$ converges for all $x \in (-R,R)$.
And now consider the following proof:
I am not sure this is correct. My doubts are the following:
(1) Is it correct to write $$ \sum_{n=1}^\infty |n a_n x^{n-1}| = \dots$$
before proving it converges?
(2) By assumption $\sum_n a_n x^n$ only converges but not necessarily absolutely. Why is it clear that $$ \sum_{n=1}^\infty |a_n t^n|$$ is finite?
Yes. $\sum x_n = \sum y_n$ here means that the sequences of terms of each are the same. In particular, it is then correct that if the second sum converges, so does the first. Similarly, $\sum x_n \leq \sum y_n$ here means that the $n$'th term of the left hand sum is smaller than the $n$'th term of the right-hand sum.
This is how convergence works for power series. Inside the interval of convergence, you converge absolutely; outside it, nothing converges at all; and on the boundary, it depends. Here by assumption $t<R$, so we're good.