Prove that if $S=\mathbb{N}$ and $\sum_{s\in S} f(s)=a$ then $\sum_{n=1}^\infty f(n) =a$ Let $f:S\rightarrow\mathbb{R}$.
Then $\sum_{s\in S} f(s)=\infty$ means For all $M\in\mathbb{R}$, there is a finite set $T\subseteq S$ such that, for all finite sets $T'\subseteq S$ that contain $T$ one has $\sum_{s\in T'} f(s)>M$
and $\sum_{s\in S} f(s)=a$ means For all $\epsilon>0$, there is a finite set $T\subseteq S$ such that, for all finite sets $T'\subseteq S$ that contain $T$, one has $|a-\sum_{s\in T'} f(s)|<\epsilon$
Prove:  If $S=\mathbb{N}$ and $\sum_{s\in S} f(s)$ converges to $a$, prove that $\sum_{n=1}^\infty f(n) =a$ (the latter meaning
the limit of the sequence of partial sums $\sum_{n=1}^N f(n)$). 
So I start my proof as:
Let $f:S\rightarrow \mathbb{R}$. Suppose $S=\mathbb{N }$ and $\sum_{s\in S} f(s)=a$. That is for all $\epsilon>0$, there is a finite set $T\subseteq S$ such that, for all finite sets $T'\subseteq S$ that contain $T$, one has $|a-\sum_{s\in T'} f(s)|<\epsilon$.
Basically just writing down what I know. Then I have to wind up here: $\lim_{N\rightarrow\infty}(\sum_{n=1}^\infty f(n))=a$. I feel like this should be very easy since $S=\mathbb{N}$. The statements $\sum_{s\in S} f(s)=a$ and $\sum_{n=1}^\infty f(n) =a$ are pretty much identical. Just the first doesn't depend upon indexing of the elements. So, how can I proceed with my proof to get where I need to go?
 A: HINT: Starting by writing out what the hypotheses mean is good. A good next step is to do the same kind of expansion of what you’re trying to prove: $\lim_{n\to\infty}\sum_{k=1}^nf(k)=a$ means that 

for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $$\left|a-\sum_{k=1}^nf(k)\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon\;.\tag{1}$$

Thus, you want to start with an arbitrary $\epsilon>0$ and use the hypothesis that $\sum_{s\in S}f(s)=a$ to find a suitable $m_\epsilon$. That hypothesis tells you that there is a finite $T_\epsilon\subseteq S=\Bbb N$ such that 
$$\left|a-\sum_{s\in T'}f(s)\right|<\epsilon\quad\text{whenever}\quad T'\text{ is finite and }T_\epsilon\subseteq T'\subseteq\Bbb N\;.\tag{2}$$
Somehow we should use $T_\epsilon$ to get $m_\epsilon$. 


*

*Prove that we can replace $T_\epsilon$ with any finite subset of $\Bbb N$ that contains $T_\epsilon$, and $(2)$ will still be true. This is the key step in the argument.


In particular, we can ‘fill in the holes’ in $T_\epsilon$ by replacing it with 
$$\{k\in\Bbb N:k\le\max T_\epsilon\}=\{1,2,\ldots,\max T_\epsilon\}\;.$$
Can you see how to use this to find an $m_\epsilon$ making $(1)$ true?
A: Consider the convergent series 
$$\sum_{s\in\mathbb{N}} f(s)$$
with sum $a$, and label this series $A$.
Now let us order the terms in series $A$ as $f(n)$, for $n=1,2,\dotsc,$ to give the series as follows:
$$\sum_{n=1}^{\infty} f(n),$$
and label this series $B$. For every partial sum $B_M=\sum_{n=1}^{M} f(n)$ of series $B$, there exists a subset of $\mathbb{N}$, say $T$, which gives a partial sum of $A$ containing all terms of $B_M$:
$$A_T=\sum_{s\in T} f(s).$$
We can now find a partial sum $B_N=\sum_{n=1}^{N} f(n)$ containing all terms from $A_T$. This gives the inequality,
$$
B_M\le A_T\le B_N.
$$
Since we know the sum of the convergent series $A$ is $a$ we also have $A_T\le a$, which, since $B_M\le A_T\le a$, implies $B$ will also be convergent with some limit $b$. Now if we let $M\rightarrow\infty$ in $B_M\le A_T\le B_N$, we get $b\le a\le b$, and so $b=a$. $\square$
