# Absolute Convergence and Supremum?

How would one prove the following?

If $\sum_{n}a_{n}$ of non-negative terms is absolutely convergent and $\sum_{n}a_{n}=a$, then $a=\sup\{S\}$, where $S$ is the set of all finite sums of values of $a_{n}$.

Is there something obvious that I'm missing? Thanks in advance for any help!

Well it is kind of obvious. Any finite sum is less than or equal to some partial sum, so no finite sum can exceed $a$, hence the sup can't exceed $a$. But the partial sums are finite sums, so the sup can't be less than $a$.

Let $S_n$ denote the partial sum $$S_n=\sum_{i=1}^na_i$$ Then the set $\,S=\{S_1,S_2,S_3,\cdots\}$

Now according to the "$\,\varepsilon\text{-}N\,$" definition of $\,\sum a_n\rightarrow a\,$, we have $$\qquad\qquad\qquad\quad\forall\varepsilon>0,\ \exists N\in\mathbb N\ \,\text{such that}\ \ n\geq N \Rightarrow\ |S_n-a|<\varepsilon\qquad\qquad(\%)$$

Next, consider two cases:

(1) Suppose that $\sup\{S\}<a$, then let $\,\varepsilon=a-\sup\{S\}>0$. According to $(\%)$,

$$\exists N\in\mathbb N\ \,\text{such that}\ \ n\geq N \Rightarrow\ |S_n-a|<\varepsilon$$ $$\Rightarrow\ S_n>a-\varepsilon=\sup\{S\}$$ which is a contradiction of $\,\sup\{S\}\,$ being the supremum

$\left.\right.$

(2) Suppose that $\sup\{S\}>a$, then let $\,\varepsilon=(\sup\{S\}-a)/2>0$. According to $(\%)$,

$$\exists N\in\mathbb N\ \,\text{such that}\ \ n\geq N \Rightarrow\ |S_n-a|<\varepsilon$$ $$\Rightarrow\ S_n<a+\varepsilon<\sup\{S\}$$ Since $\{S_n\}$ is strictly increasing and $\,S_n<a+\varepsilon<\sup\{S\}\,$, hence $\,\sup\{S\}\,$ is not the $\textbf{least}$ upper bound, which causes contradiction.

$\left.\right.$

As a result, $\sup\{S\}=a$