Sum of convergent power series with increasing degrees is convergent.

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I'm contemplating Exercise 1.1.3 from Greuel, Lossen and Shustin's Introduction to Singularities and Deformations. It states

Let $(f_k)_{k \in \mathbb N}$ be a sequence of convergent power series in $\mathbb C\{x_1, \dotsc, x_n\}$ with $\operatorname{ord}(f_{k+1}) > \operatorname{ord}(f_k)$ for all $k$. Show that $\sum_{k \in \mathbb N} f_k$ is a well-defined convergent power series.

For a power series $f = \sum_\alpha c_\alpha x^\alpha$, the order is defined as $$ \operatorname{ord}(f) = \min \{\, |\alpha| : c_\alpha \neq 0 \,\}.$$

Clearly, $g = \sum_k f_k$ is a well-defined formal power series, since for every multiindex $\alpha \in \mathbb N_0^k$, the term $g_\alpha$ only involves such $f_k$ with $\operatorname{ord}(f_k) \leq |\alpha|$ where $| \alpha| = \alpha_1 + \dotsb + \alpha_n$, and those are finitely many by assumption. But I don't know how to show that $g$ is convergent.

However, I'm not sure if this is even true? For example consider the power series $$ f_k = \sum_{n = k}^\infty k^n x^n.$$ This is convergent by the ratio test with radius of convergence $r_k = \frac{1}{k}$. Since $\operatorname{ord}(f_k) = k$, the assumption $\operatorname{ord}(f_{k+1}) > \operatorname{ord}(f_{k})$ is satisfied. But for each $\epsilon > 0$, all but finitely many $f_k$ are divergent in a disc of radius $\epsilon$, so how can we even expect $\sum_k f_k$ to be convergent?

More precisely, the power series here is $$ g =\sum_{k \in \mathbb N} f_k = \sum_{k \in \mathbb N} \sum_{n \geq k} k^n x^n = \sum_{n \in \mathbb N} \left( \sum_{k \leq n} k^n \right) x^n. $$ So by the formel of Cauchy-Hadamard, it has radius of convergence $$ r = \lim_{n \to \infty} \frac{1}{\sqrt[n]{\sum_{k \leq n} k^n}} \leq \lim_{n \to \infty} \frac{1}{\sqrt[n]{n^n}} = \lim_{n \to \infty} \frac{1}{n} = 0. $$