When does $\lim_{s\to 1_-} (1-s)\sum_{n=0}^\infty a_ns^n$ exist? Let $a=\{a_n\}_{n\geq 0}$ be a sequence of positive real numbers with $a_n\leq 1$, for all $n$, and observe that, for
any real number $s\in [0,1)$, one has that
$$
  \sum_{n=0}^\infty a_ns^n \leq \sum_{n=0}^\infty s^n = \frac 1 {1-s}.
  $$
Therefore the function $f_a$ defined by
$$
  f_a(s) = (1-s)\sum_{n=0}^\infty a_ns^n, \quad \forall s\in [0,1),
  $$
is bounded by 1.  My main interest is to discuss the existence of the limit of $f_a$ as $s$ tends to 1 from the left,
namely
$$
  \lim_{s\to 1_-} (1-s)\sum_{n=0}^\infty a_ns^n.
  $$
Plugging in most naive examples of sequences $\{a_n\}$, one would be tempted to conjecture that the limit always
exists, but I've been shown examples where it doesn't.
My gut impression is that this problem belongs to some well established theory, perhaps related to analytic number
theory and this is precisely what brings me here.
Question.  Is there a particular point of view in Math from where the existence of the above limit has been
discussed?
 A: The answer is negative.
I set $a_n = 1_B(n)$, where $B$ is the union of all intervals of the form $[(2k)!,(2k+1)![$.
Let $N_s$ be a random variable with geometric distribution on $\{0,1,\ldots\}$ with parameter $1-s$. For every $n \ge 0$, $P[N_s=n] = (1-s)s^n$. Thus the quantity studied is $E[1_B(N_s)] = P[N_s \in B]$.
When $s \to 1$, the random variable $(1-s)N_s$ converges in distribution to an exponential $X$ random variable with parameter $1$.
Let $n_k = \lfloor k!\sqrt{k+1} \rfloor$ and $s_k=1-1/n_k$. As $k \to +\infty$, we have $N_{s_k}/n_k \to X$ in distribution so $P[k! \le N_{s_k} < (k+1)!] \to 1$ since $k!<<n_k<<(k+1)!$.
The event $P[k! \le N_{s_k} < (k+1)!]$ is contained in the event
$[N_{s_k} \in B]$ if $k$ is even, and in $[N_{s_k} \notin B]$ if $k$ is odd. Hence, $P[N_{s_k} \in B]$ tends to $1$ for even $1$, and to $0$ for odd $n$.
A: Using summation by parts, we have
$$
f_a(s)=\sum_{n\ge0}a_n(s^n-s^{n+1})=a_0+\sum_{n\ge1}(a_n-a_{n-1})s^n
$$
After this transformation, apply Abelian theorem to the remaining power series.
