# What is a formal definition of series?

Is there a formal definition for series? For example, cardinal sum has a formal definition such that $\sum a_i$ = $\bigcup a_i$. Is there any clear definition for series of real or complex number?

The definition on my book is the sum of $a_0 + ... + a_n$. This seems very intuitive to me so i don't like it..

I tried to define series such that $\gamma(0) = a_0$ and $\gamma(n+1) = f_n(\gamma(n))$ where $f_n(x) = x+ a_n$

Since $f_n ≠ f_{n+1}$, i can't apply finite recurssion theorem. However it seems obvious that $\gamma$ is a function and unique. How do i prove the existence and uniqueness of $\gamma$?

I have proved that "If A is a set, 'c' a fixed point in A and $f_n : A →A$ a function for every $n\in \mathbb{N}$, then there exists a unique function $\gamma : \mathbb{N} →A$ such that $\gamma(0) = c$ and $\gamma(n+1) = f_n(\gamma(n))$. This is a bit generalized form of original finite recursion theorem. Let $\alpha$ be a sequence. Let $f_n(x)=x+\alpha(n+1) : F→F$. (F denotes an arbitrary field here) Then by above theorem, we can construct $\gamma$. It can be easily checked that $\gamma(n)$ is a summation of $a_0,...,a_n$.

By the way, i've never said series is a 'finite summation'. Of course, series is a sequence..

-
Sure; look in any real analysis textbook. – Qiaochu Yuan Jun 24 '12 at 7:32
@Qiaochu given a sequence {$a_n$}, $\sum a_n$ denotes $a_0$ + ... $a_n$, p59. PMA -Rudin – Katlus Jun 24 '12 at 7:40
Well, that's because there isn't such a function (consider what happens if $\gamma(n) = \gamma(m)$ but $a_{n+1} \neq a_{m+1}$). – Qiaochu Yuan Jun 24 '12 at 7:52
Once again: to accept an answer less than 45 minutes after the question was asked has some questionable consequences. – Did Jun 24 '12 at 8:21
@Katlus Please notice that $a_0+\ldots +a_n$ is not a series. You should probably write $\{a_0+\ldots + a_n\}_{n \geq 1}$ or something alike. – Siminore Jun 24 '12 at 8:30

Take any sequence $(a_0,a_0,\ldots)$ of real numbers. Define a function $r$ on finite sequences of real numbers such that if $s$ has length $n-1>0$, $r(s)$ is the last value of $s$ plus $a_n$. If the length of $s$ is $0$ (which we allow), then $r(s)=a_0$. The (countable) recursion theorem gives you a unique sequence $(b_0,b_1\ldots)$ such that $b_{n+1}=b_n+a_{n+1}$ and that defines your series. Note that the recursion is allowed to depend on all finite sequence of values.

There are two things you might think are not rigorous here: The "length" and the "last value". A finite sequence is a function with domain being a finite ordinal. This ordinal is the length, which is therefore well-defined. the elements of an ordinal are again ordinals and the value at the largest such ordinal in the domain is the "last value".

-
Sorry that i changed my question while you were posting it .. Would you please tell me how do i define such $b$? The finite recurssion theorem i know has $f$ doesn't depend on its 'input' value..(described in my edited question) – Katlus Jun 24 '12 at 8:13
You cannot use the recursion theorem you mentioned, which is why use a stronger version in which one takes finite sequences as inputs. It is a special version of the general recursion theorem you find in set theory books. – Michael Greinecker Jun 24 '12 at 8:17
I have proved that "If A is a set, 'c' a fixed point in A and $f_n : A →A$ a function for every $n\in \mathbb{N}$, then there exists a unique function $\gamma : \mathbb{N} →A$ such that $\gamma(0) = c$ and $\gamma(n+1) = f_n(\gamma(n))$. – Katlus Jun 24 '12 at 12:08
which is a little generalized form of original finite recursion theorem. With this I have defined series as a sequence and it works fine. Thanks for your help :) – Katlus Jun 24 '12 at 12:10

$$\sum_{i=1}^2a_i=a_1+a_2$$

$$\sum_{i=1}^{n+1}a_i=\sum_{i=1}^na_i+a_{n+1}$$

$$\sum_{i=1}^{\infty}a_i=\lim_{n\to\infty}\sum_{i=1}^na_i$$ if the limit exists.

-
I do not like this definition. I believe that a series is a sequence. You write that a series is the limit of a sequence. This might generate some confusion, because you should not even speak of the series $\sum_n (-1)^n$. – Siminore Jun 24 '12 at 8:24
I am quite happy to speak of the series $\sum_n(-1)^n$; I speak of it as a divergent series. But if you don't like that, define $\sum^{\infty}a_i$ to be the sequence $(\sum^1a_i,\sum^2a_i,\sum^3a_i,\dots)$. – Gerry Myerson Jun 24 '12 at 9:23
Well, we should probably agree on the most economic definition. – Siminore Jun 24 '12 at 9:47
@Siminore, the most economic definition is "six". – Gerry Myerson Jun 24 '12 at 12:20
I can't understand the last comment, sorry. – Siminore Jun 24 '12 at 12:22

To answer the question in the title, a series is a pair of sequences $(x_n)_{n\geqslant0}$ and $(S_n)_{n\geqslant0}$ such that $S_n=\sum\limits_{k=0}^nx_k$ for every $n\geqslant0$ (with the obvious modifications for sequences and series indexed by $n\geqslant i$ with $i\ne0$).

-
I think this is a formal but correct definition. – Siminore Jun 24 '12 at 8:25
@Siminore: Yes, formal (whatever that means) and correct. – Did Jun 24 '12 at 8:27
Of course :-) I wrote that because the OP seems to prefer a rather "colloquial" definition. By the way, His/her textbook states that a series is a finite sum! – Siminore Jun 24 '12 at 8:28
@Siminore: Yep, you are right that there are definitely some serious misconceptions at work here. Unfortunately, apart from that, I am not even sure of what the question is really... – Did Jun 24 '12 at 8:36