This may be silly but I need to be set straight.

I've spent the last 20 pages of the book learning why if it works for finite things it doesn't necessarily work for countably infinite, of course if something works for countably infinite it works for finite amount.

Anyway I know $\sum^n_{i=1}a_i\le b$ say for all n. How can I jump to $\sum^\infty_{i=1}a_i\le b$?

I want to say limits. if $\lim_{n\rightarrow\infty}(\sum^n_{i=1}a_i)\le b$ then

$\forall\epsilon>0\exists N\in\mathbb{N}:n>N\implies \sum^n_{i=1}a_i-b<\epsilon$

Which is trivially seen to be true.

Is it because that $b$ is not a function of $n$ this works, b works for all n being the critical thing that means this works for countably infinite, rather than just finite?

(I think so, but without a book holding my hand I get scared!)

  • $\begingroup$ Yes, the fact that b is independent of n is essential. $\endgroup$ – mathematician Jan 19 '15 at 15:29
  • $\begingroup$ @mathematician That's not true. If $s_n\leqslant t_n$ for all but finitely many $n$, then $\lim s_n\leqslant \lim t_n$ (provided the limits exist). In this case $t_n=b$ for every $n$ and $s_n=\sum^na_i$. $\endgroup$ – Pedro Tamaroff Jan 19 '15 at 15:30
  • $\begingroup$ If $\sum\limits_{i=1}^{\infty} a_i$ exists, then your statement is true. It might be easier to prove the contrapositive. Take $t=\sum\limits_{1}^{\infty} a_i$. If $t>b$, then take $\varepsilon = ({t-b})/{2}$ and arrive at a contradiction. $\endgroup$ – d80d2729a352b1366139fc119d3345 Jan 19 '15 at 15:34
  • $\begingroup$ @PedroTamaroff what? In this case s_n <= b FOR ALL n $\endgroup$ – Alec Teal Jan 20 '15 at 7:50
  • $\begingroup$ Minor nitpick. You know $a$ is true and you want to prove $b$. So you show $b \implies a$ so $b$ being true is consistent. Which doesn't actually prove $b$ is true. Just that b might be true. Prove if $a_n \le b$ for all $n$ and $\lim a_n = c$ then $c \le b$. $\endgroup$ – fleablood Jul 22 '17 at 20:25

Well, to go to the definitions.

If $\sum_{i=1}^n a_i \le b$ for all $i$, then sequence $\{\sum_{i=1}^n a_i| n \in \mathbb N\}$ is bounded above by $b$. So as the reals have the least upper bound property $\beta= \sup \{\sum_{i=1}^n a_i| n \in \mathbb N\}\le b$ exists.

We actually have a problem in that we don't actually know that $\lim_{n\rightarrow \infty} \sum_{i=1}^n a_i$ exists. It's possible $a_i =(-1)^i$ and $\sum_{i=1}^{n} a_i = \{-1|0\} \le 0$ but $\sum_{i=1}^{\infty} a_i$ does not exist.

However if we know that $\sum a_i$ is increasing we are okay. (For instance if $a_i \ge 0$).

We know that if $\{c_n\}$ is a sequence and $c_i \le c_{i+1}$ and $c_i \le b$ then $\lim_{n\rightarrow \infty} c_i = \beta = \sup \{c_i\}$.

That follows by definition. For any $\epsilon > 0$ then $\beta - \epsilon < \beta$ is no an upper bound (as $\beta$ is least upper bound.) So there is an $M$ so that $\beta - \epsilon < c_M \le \beta$. So for all $n \ge M$, $c_n \ge c_M > \beta - \epsilon$. So by definition $\lim_{n\rightarrow \infty} c_i = \beta = \sup \{c_i\}$

And we are done:

The sequence $\{\sum_{i=1}^n a_i| n \in \mathbb N\}$ and is bounded above so $\beta =\sup \{\sum_{i=1}^n a_i| n \in \mathbb N\}$ exists and as the sequence is increasing $\sum_{i=1}^{\infty}a_i := $(by definition) $\lim_{n\rightarrow \infty} \sum_{i=1}^{n}a_i = \beta= \sup \{\sum_{i=1}^n a_i| n \in \mathbb N\}\le b$.


One way to describe this example is to say that the set $B=(-\infty,b]=\{c\in \mathbb R: c\leq b\}$ is a closed set, or to say it is a sequentially closed set: If $(c_n)_{n\in \mathbb N}$ is a sequence of members of $B$ that converges to c, then $c\in B.$

So if $c_n=\sum_{i=1}^na_i\in B$ for each $n,$ and if $c=\lim_{n\to \infty}c_n$ exists, then $\sum_{i=1}^{\infty}a_i=\lim_{n\to \infty}c_n=c\in B.$

One thing that doesn't work at infinity is that if $c_n<b$ for all $n$, and if $c=\lim_{n\to \infty}c_n$ exists, we cannot conclude that $c<b,$ but only that $c\leq b.$ For example let $b=1$ and $c_n=1-2^{-n}.$ Then $c=\lim_{n\to \infty} c_n=1=b.$


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