Proof of Sums of Series: a Convergent Series Has a Real Sum, and a Bound on All Partial Sums Bounds the Sum in [0,∞]

Proof of (i). Let $(s_{n})$ be the sequence of partial sums of $(a_{k})$. That the series converges means, by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii), that $(s_{n})$ is convergent, and its sum is then the limit of $(s_{n})$:

$$\sum_{k=1}^{\infty}a_{k}=\lim_{n\to\infty}s_{n}.$$

By The Limit of a Convergent Sequence (i), the limit of a convergent sequence is the real number to which it converges. Hence $\sum_{k=1}^{\infty}a_{k}\in\mathbb{R}$.

Proof of (ii). Let $A=\{s_{n}:n\in\mathbb{N}\}$ (a subset of $[0,\infty]\subseteq[-\infty,\infty]$, by Sum of a Sequence in [0,∞] (i) and The Extended Real Line, Its Order, and [0,∞] (ii)). By Sum of a Sequence in [0,∞] (ii),

$$\sum_{k\in\mathbb{N}}a_{k}=\sup A,$$

the supremum formed in $[-\infty,\infty]$ with $\le$ (it exists by Order and Suprema in the Extended Real Line (iii)). Since $s_{n}\le t$ for every $n\in\mathbb{N}$, $t$ is an upper bound of $A$. By Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv), $\sup A$ is a least element of the set of upper bounds of $A$, so $\sup A\le t$, that is, $\sum_{k\in\mathbb{N}}a_{k}\le t$.