TITLE: Why is an infinite series not considered an infinite sum of terms?
QUESTION [13 upvotes]: According to, for example, this excellent page on beginner calculus, an infinite series is NOT an infinite sum of terms.* I'm not even sure what that's asserting.
I think an infinite series is an infinite number of terms that are added together. I also think you can call it the limit of its partial sum $S_n$ as $n \to \infty$ but I'm not sure if or how those two ideas conflict.
Does the infinite sum of terms 'equal' the limit of a series' partial sum, and if not why not?

*To quote: "We do have to be careful with this however. This implies that an infinite series is just an infinite sum of terms and as we’ll see in the next section this is not really true." I looked at the next section and I still don't get what they're saying.

REPLY [12 votes]: Series do not enjoy all the nice properties usual finite sums do. For example:
Altering parenthesis may alter the sum. We know that  $$(1-1)+(1-1)+\cdots=0+0+\cdots=0$$ but   $$1+(-1+1)+(-1+1)+\cdots=1+0+0+\cdots=1$$
Thus, introducing parenthesis alters the sum. And deleting them:
$$1-1+1-1+1-1+1-+\cdots$$
gives us a divergent sum!
Two convergent series may produce a divergent series upon multiplication Take $$S=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{\sqrt n}$$
One can show this converges by use of Leibniz's Test. But the Cauchy product of $S$ with itself diverges, because the harmonic series diverges.
The sum of rational numbers may yield an irrational number Take, as an example $$\sum_{n\geq 1} \frac{1}{n^2}=\frac{\pi^2}6$$$$\sum_{n\geq 1} \frac{(-1)^{n-1}}n=\log 2$$
The order in which we sum terms matters For example, if we take the Harmonic series, we can always take positive and negative terms to our liking to make the sum any number we want. This is a particular case of a theorem by Riemann