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I'm starting to read Real Analysis. Consider below:

$$ \lim_{n\to\infty} \sum_{i=1}^{n}x_i = \sum_{i=1}^{\infty}x_i $$

This seems obvious and intuitive, but:

Q1 - How do you prove this (rigorously, at the level of Baby Rudin)?

Q2 - How do you formally / rigorusly define, $\sum_{i=1}^{\infty}x_i$? What if the summation is not defined / doesn't exist (i.e. is divergent). For e.g. $\sum_{i=1}^{\infty}i$ ?

Thanks

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    $\begingroup$ The quantity on the left is definition of the quantity on the right... $\endgroup$ Commented Oct 20 at 12:26
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    $\begingroup$ You are correct, I suppose, in thinking that there is a small abuse of notation here...in that the right hand is defined by the left in the event that the limit exists. Of course, the limit need not exist. $\endgroup$ Commented Oct 20 at 12:34
  • $\begingroup$ Thanks got it, and if the limit doesn't exist then do we say that right hand summation is "not defined" - I think that is how it should be else it will lead to some absurd results like in the case of division by 0 $\endgroup$ Commented Oct 20 at 12:41
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    $\begingroup$ Yes. Just as one might write a rational function like $f(x)=\frac 1{x^2-1}$ without troubling to mention that this fails to define a value if $x=\pm 1$. Writing the right hand sum comes with the implicit comment "if that limit exists". $\endgroup$ Commented Oct 20 at 13:08

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For any sequence $(a_n)_n \subset \mathbb R$, we say that $a \in \mathbb R$ is the limit of $a_n$ if $\forall \varepsilon > 0, \exists N \in \mathbb N $ such that $\forall n \ge N,$ we find $$|a_n - a| \le \varepsilon.$$ In this case, we denote (this is just a notation): $$\lim_{n \to \infty} a_n := a.$$ We say that a sequence diverges if the does not converge, that is $$\forall a \in \mathbb R, \exists \varepsilon > 0, \forall N \in \mathbb N, \exists n \ge N: |a_n - a| > \varepsilon.$$ Similarly, we say the series of $a_n$ converges/diverges if the sequence of partial sums $$S_n = \sum_{k = 1}^n a_k$$ converges/diverges in the sense described before.

In the case of a divergent series, the notation $$\sum_{i = 1}^\infty a_n$$ does not make sense in $\mathbb R$ since the limit of $S_n$ does not exist.

However, in the example you gave, notice that $$S_n = \sum_{i = 1}^n i \to +\infty \quad \text{as }n \to \infty.$$ Therefore, even though $\lim_n S_n$ is not a real number, we sometime denote $$\sum_{i = 1}^\infty i = +\infty,$$ to suggest that the limit of $S_n$ is in $\mathbb R \cup\{+\infty, -\infty\}$.

If you consider the partial sums of $$S_n = \sum_{k = 1}^n(-1)^k$$ the notation $$\sum_k^\infty (-1)^{k}$$ is simply nonsensical in $\mathbb R$ or $\mathbb R \cup\{+\infty, - \infty\}$.

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As mentioned in the comments, what you stated is the usual definition of a series, so to answer your questions:

A1: There is nothing to prove. By definition, $\sum_{i=1}^\infty x_i$ is the limit of the sequence of partial sums: $$\sum_{i=1}^\infty x_i=\lim_{n\to\infty}\sum_{i=1}^n x_i$$
A2: See A1. The definition makes since even if the series diverges. I am not sure what do you mean by "doesn't exist".

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  • $\begingroup$ This should be a comment, not an answer. $\endgroup$ Commented Oct 20 at 12:59

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