Absolute Convergence and Supremum?

Question

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!

Answer 1

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 $a$.

Answer 2

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

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(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.

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