[proofplan] The key structural feature is that every term $u_n s^n$ is non-negative, which allows us to exchange the limit and the sum without any domination hypothesis. We establish the two inequalities separately. For the lower bound, we truncate the series at a finite index $N$, take $\liminf$ as $s \to 1^-$, and then supremise over $N$. For the upper bound, we use the pointwise estimate $u_n s^n \leq u_n$ for $s \in (0,1)$ to bound $U(s)$ above by the full series term-by-term. Combining the two inequalities shows that $\lim_{s \to 1^-} U(s)$ exists in $[0, \infty]$ and equals $\sum_{n=0}^\infty u_n$. [/proofplan] [step:Bound $\liminf_{s \to 1^-} U(s)$ from below by truncating at a finite index $N$] Fix $N \in \mathbb{N} \cup \{0\}$. Since $u_n \geq 0$ for all $n$, every term in the series defining $U(s)$ is non-negative. Retaining only the first $N + 1$ terms and discarding the (non-negative) remainder gives the inequality \begin{align*} U(s) = \sum_{n=0}^{\infty} u_n s^n \geq \sum_{n=0}^{N} u_n s^n \end{align*} for all $s \in (0,1)$. The right-hand side is a polynomial in $s$ — in particular a finite sum of continuous functions — so \begin{align*} \lim_{s \to 1^-} \sum_{n=0}^{N} u_n s^n = \sum_{n=0}^{N} u_n \cdot 1^n = \sum_{n=0}^{N} u_n. \end{align*} Taking $\liminf_{s \to 1^-}$ of both sides of the inequality $U(s) \geq \sum_{n=0}^{N} u_n s^n$ and using the fact that the right-hand side has an actual limit, we obtain \begin{align*} \liminf_{s \to 1^-} U(s) \geq \sum_{n=0}^{N} u_n. \end{align*} The left-hand side does not depend on $N$. Since this inequality holds for every $N \in \mathbb{N} \cup \{0\}$, we take the supremum over all $N$ on the right-hand side. By definition, $\sup_{N \geq 0} \sum_{n=0}^{N} u_n = \sum_{n=0}^{\infty} u_n$ (as the supremum of partial sums of a non-negative series). Therefore \begin{align*} \liminf_{s \to 1^-} U(s) \geq \sum_{n=0}^{\infty} u_n. \end{align*} [guided] The strategy for the lower bound is to replace the infinite series $U(s)$ by something whose limit as $s \to 1^-$ we can compute exactly — namely a finite partial sum. Why does truncation give a lower bound? Because every omitted term $u_n s^n$ is non-negative (both $u_n \geq 0$ and $s^n > 0$ for $s \in (0,1)$), so dropping terms can only decrease the sum. Fix $N \in \mathbb{N} \cup \{0\}$. Retaining only the first $N + 1$ terms of the series and discarding the non-negative tail: \begin{align*} U(s) = \sum_{n=0}^{\infty} u_n s^n \geq \sum_{n=0}^{N} u_n s^n. \end{align*} The right-hand side is a polynomial in $s$, so we may evaluate its limit as $s \to 1^-$ term by term (a finite sum of continuous functions): \begin{align*} \lim_{s \to 1^-} \sum_{n=0}^{N} u_n s^n = \sum_{n=0}^{N} u_n \cdot 1^n = \sum_{n=0}^{N} u_n. \end{align*} Now we take $\liminf_{s \to 1^-}$ of the inequality $U(s) \geq \sum_{n=0}^{N} u_n s^n$. The monotonicity property of $\liminf$ (if $f(s) \geq g(s)$ for all $s$, then $\liminf f \geq \liminf g$) gives \begin{align*} \liminf_{s \to 1^-} U(s) \geq \liminf_{s \to 1^-} \sum_{n=0}^{N} u_n s^n = \sum_{n=0}^{N} u_n, \end{align*} where the equality on the right uses the fact that $\liminf$ equals the actual limit when the limit exists. The left-hand side is a fixed quantity — it does not depend on $N$. Since the inequality holds for every $N$, we may take $\sup_{N \geq 0}$ on the right-hand side: \begin{align*} \liminf_{s \to 1^-} U(s) \geq \sup_{N \geq 0} \sum_{n=0}^{N} u_n = \sum_{n=0}^{\infty} u_n, \end{align*} where the last equality is the definition of an infinite series with non-negative terms as the supremum of its partial sums. [/guided] [/step] [step:Bound $\limsup_{s \to 1^-} U(s)$ from above by the term-by-term comparison $u_n s^n \leq u_n$] For each $n \geq 0$ and each $s \in (0,1)$, we have $0 < s < 1$, so $s^n \leq 1$. Since $u_n \geq 0$, this gives the pointwise bound \begin{align*} u_n s^n \leq u_n \quad \text{for all } n \geq 0 \text{ and all } s \in (0,1). \end{align*} Summing over $n \geq 0$ (both sides are series of non-negative terms, so the inequality is preserved under summation): \begin{align*} U(s) = \sum_{n=0}^{\infty} u_n s^n \leq \sum_{n=0}^{\infty} u_n. \end{align*} The right-hand side is a constant (independent of $s$), so taking $\limsup_{s \to 1^-}$: \begin{align*} \limsup_{s \to 1^-} U(s) \leq \sum_{n=0}^{\infty} u_n. \end{align*} [guided] For the reverse inequality, we need an upper bound on $U(s)$ that does not depend on $s$. The natural candidate is the full series $\sum_{n=0}^{\infty} u_n$ itself, which we obtain by replacing each factor $s^n$ by its upper bound $1$. For each $n \geq 0$ and $s \in (0,1)$: since $0 < s < 1$, we have $s^n \leq s^0 = 1$ (powers of a number in $(0,1)$ are at most $1$). Multiplying by $u_n \geq 0$ preserves the inequality: \begin{align*} u_n s^n \leq u_n \quad \text{for all } n \geq 0, \; s \in (0,1). \end{align*} We now sum both sides over $n = 0, 1, 2, \dots$. Both are series of non-negative terms, and summing non-negative terms preserves the termwise inequality (if $a_n \leq b_n$ for all $n$ with $a_n, b_n \geq 0$, then $\sum a_n \leq \sum b_n$ in $[0, \infty]$): \begin{align*} U(s) = \sum_{n=0}^{\infty} u_n s^n \leq \sum_{n=0}^{\infty} u_n. \end{align*} The upper bound $\sum_{n=0}^{\infty} u_n$ is a constant — it does not depend on $s$. Since $U(s) \leq C$ for all $s \in (0,1)$ where $C = \sum_{n=0}^{\infty} u_n \in [0, \infty]$ is fixed, we have \begin{align*} \limsup_{s \to 1^-} U(s) \leq C = \sum_{n=0}^{\infty} u_n. \end{align*} Note that this bound is valid whether $\sum u_n$ is finite or $+\infty$ (in the latter case the inequality $\limsup U(s) \leq +\infty$ is vacuously true, but the lower bound from the previous step will force equality). [/guided] [/step] [step:Combine the two bounds to conclude $\lim_{s \to 1^-} U(s) = \sum_{n=0}^{\infty} u_n$] From the first step, $\liminf_{s \to 1^-} U(s) \geq \sum_{n=0}^{\infty} u_n$. From the second step, $\limsup_{s \to 1^-} U(s) \leq \sum_{n=0}^{\infty} u_n$. Since $\liminf \leq \limsup$ always holds, we obtain the chain of inequalities \begin{align*} \sum_{n=0}^{\infty} u_n \leq \liminf_{s \to 1^-} U(s) \leq \limsup_{s \to 1^-} U(s) \leq \sum_{n=0}^{\infty} u_n. \end{align*} All four quantities are therefore equal. In particular, $\liminf_{s \to 1^-} U(s) = \limsup_{s \to 1^-} U(s)$, which means $\lim_{s \to 1^-} U(s)$ exists in $[0, \infty]$ and \begin{align*} \lim_{s \to 1^-} U(s) = \sum_{n=0}^{\infty} u_n. \end{align*} If $\sum_{n=0}^{\infty} u_n < \infty$, then $\lim_{s \to 1^-} U(s) < \infty$; conversely, if $\lim_{s \to 1^-} U(s) < \infty$, then $\sum_{n=0}^{\infty} u_n < \infty$. In either case, both sides are equal. If both are $+\infty$, the equality $+\infty = +\infty$ still holds. $\blacksquare$ [guided] We have established two inequalities: - **Lower bound** (from the first step): $\liminf_{s \to 1^-} U(s) \geq \sum_{n=0}^{\infty} u_n$, obtained by truncating the series and taking the supremum over truncation indices. - **Upper bound** (from the second step): $\limsup_{s \to 1^-} U(s) \leq \sum_{n=0}^{\infty} u_n$, obtained from the term-by-term comparison $s^n \leq 1$. The general inequality $\liminf \leq \limsup$ (which holds for any function) combines with these two bounds to give \begin{align*} \sum_{n=0}^{\infty} u_n \leq \liminf_{s \to 1^-} U(s) \leq \limsup_{s \to 1^-} U(s) \leq \sum_{n=0}^{\infty} u_n. \end{align*} Since the leftmost and rightmost quantities are equal, all intermediate quantities must agree. The equality of $\liminf$ and $\limsup$ is precisely the condition for the limit to exist, so $\lim_{s \to 1^-} U(s)$ exists in $[0, \infty]$ and equals $\sum_{n=0}^{\infty} u_n$. For the "in particular" statement: if $\sum_{n=0}^{\infty} u_n < \infty$, the equality forces $\lim_{s \to 1^-} U(s) = \sum_{n=0}^{\infty} u_n < \infty$. Conversely, if $\lim_{s \to 1^-} U(s) < \infty$, the same equality forces $\sum_{n=0}^{\infty} u_n < \infty$. Thus finiteness of either side implies finiteness (and equality) of the other. When both sides are $+\infty$, the statement $+\infty = +\infty$ holds as an equality in $[0, \infty]$. $\blacksquare$ [/guided] [/step]