Ok, this is probably a very silly question, and I'm probably overcomplicating the issue, but here it goes:
I'm making questions on power series, and it asks me to find the convergence radius of the series
$$\sum_{n=1}^\infty z^{n!}$$
Of course, usually a power series is given in the form $$\sum_{n=1}^\infty a_n z^n$$
and we see that we here have to take $$a_n = \begin{cases}1 \quad n = k! \quad \mathrm{for \ some \ k}\\ 0 \quad \mathrm{else}\end{cases}$$
From this, it is easy to see that $\limsup_{n\to \infty} |a_n|^{1/n} = 1$ as we have a subsequence converging to $1$ and the limsup can't become greater than that, so $1$ is the largest limit of a subsequence. Hence, the convergence radius is $1$.
My concern: Shouldn't we formally show that
$$\sum_{n=1}^\infty z^{n!}= \sum_{n=1}^\infty a_n z^n$$
for my choice of $(a_n)_n$? In the left sum, we sum over 'less' terms that the right one, while in the right sum we sum over 'more' terms, but these are zero so intuitively the terms are not affected.
Write $(c_n)_n$ for the left partial sums and $(d_n)_n$ for the right partial sums. Then we see that $(d_n)_n$ has the form $(c_1,\underbrace{c_1, \dots, c_1}_{k_1 terms}, c_2, \underbrace{c_2, \dots, c_2}_{k_2 terms}, \dots)$ and it suffices to show that
$(c_n)_n$ converges if and only if $(d_n)_n$ converges and in this case the limits coincide
If $(d_n)_n$ converges, then $(c_n)_n$ converges to the same limit because it contains $(c_n)_n$ as a subsequence.
Conversely, assume $(c_n)_n$ converges to $c$. Let $\epsilon > 0$ and choose $n_0$ such that $d(c_n, c) < \epsilon$ whenever $n \geq n_0$. If $n \geq n_0 + k_1 + \dots + k_{n_0}$, then $d_n\in \{c_{n_0}, c_{n_0+1}, \dots\}$ and it follows that $d(d_n, c) < \epsilon$. Hence $d_n \to c$, proving the statement. $\quad \square$.
Any thoughts about it? Am I overthinking this?
You don't need to formally show it. The $a_n$ are exactly as you said. We just take a limsup of a very "sparse" sequence (most terms are $0$) to determine the radius of convergence. The argument about there being a constant subsequence of $1$'s and thus the same holds for the root-sequence, is correct.