Given an infinite sequence (a1, a2, a3, ...), a series is informally the form of adding all those terms together: a1 + a2 + a3 + ···. To emphasize that there are an infinite number of terms, a series is often called an infinite series.
值得注意的是等式右邊並不是左邊的和，只是左邊的縮寫形式。
because when you start from adding up the first two terms of the infinite sequence, and then add the third term, the 4-th term, ..., no matter how much time you spend on adding these terms, you always end up adding up only a finite number of terms , thus you couldn't add up an infinite number of terms, so cannot compute their sum by adding one term after another.
An easy way that an infinite series has a sum is if all the $a_n$ are zero for n sufficiently large. Such a series can be identified with a finite sum, so it is only infinite in a trivial sense.
Working out the properties of the series that has a sum even if infinitely many terms are non-zero is the essence of the study of series. Consider the example
It is possible to "visualize" it has sum on the real number line: we can imagine a line of length 2, with successive segments marked off of lengths 1, ½, ¼, etc. There is always room to mark the next segment, because the amount of line remaining is always the same as the last segment marked: when we have marked off ½, we still have a piece of length ½ unmarked, so we can certainly mark the next ¼. This argument does not prove that the sum is equal to 2 (although it is), but it does prove that it is at most 2. In other words, the series has an upper bound. As for proving the series is equal to 2, we choose $$a_n=1+\frac12+\frac14+\frac18+\frac{1}{16}+\cdots+\frac{1}{2^{n-1}}+\frac{1}{2^n}$$ and $b_n=2$, then $$a_n<1+\frac12+\frac14+\frac18+\frac{1}{16}+\cdots+\frac{1}{2^{n-1}}+\frac{1}{2^n}+\cdots\leq b_n$$ holds for every nature number $n$ and $\lim _{n\rightarrow \infty }\left( b_{n}-a_{n}\right)=\lim _{n\rightarrow \infty }\frac{1}{2^n} =0$, according to the nested intervals theorem the intersection of all the $[a_n,b_n]$ contains exactly one real number, since 2 is an element of each of these intervals, $1+\frac12+\frac14+\frac18+\frac{1}{16}+\cdots+\frac{1}{2^{n-1}}+\frac{1}{2^n}+\cdots = 2$, this proved the sum of the series is 2.
It is also possible to prove $$1 - {1 \over 2} + {1 \over 3} - {1 \over 4} + {1 \over 5} - \cdots =\sum_{n=1}^\infty {\left(-1\right)^{n-1} \over n}=\ln(2)$$ using the nested intervals theorem by choosing $a_k=\sum_{n=1}^{2k} {\left(-1\right)^{n-1} \over n}$ and $b_k=\sum_{n=1}^{2k+1} {\left(-1\right)^{n-1} \over n}$ for all natural numbers $k$.
While a more general method to get the sum of a series is by taking limit.
As you see, we defined the sum of a infinite series, this result seems not that naturally like 2 + 2 is computed out equal to 4，so is the definition give us the true sum of the infinite series? $\lim _{n\rightarrow \infty }S_{n}$ has a meaning that the number of the first n terms added up increases indefinitely, this is equivalent to $$a_{1}+a_{2}+a_{3}+\cdots $$, thus defining the sum of a series as the limit of the sequence of its partial sums is intuitively plausible.
Given the definition gives the true sum of the infinite series, the statement that 0.999… = 1 can itself be interpreted and proven as:
${\displaystyle 0.999\ldots =\lim _{n\to \infty }0.\underbrace {99\ldots 9} _{n}=\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}=\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)=1-\lim _{n\to \infty }{\frac {1}{10^{n}}}=1\,-\,0=1.\,}$
批注：一開始convergent和divergent是對一個sequence來說的，定義如下
但怎么能說級數convergent和divergent了呢？級數，根據上面的定義不就是一個數列的無窮多項依次加起來的一個和式嗎？對於一個和式能說convergent和divergent嗎？我看不如說一個級數has a sum or not，然后說其部分和組成的數列convergent和divergent似乎比較合適！說一個series converges to a limit L不如說這個series =L。
quoted from http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-convergence-2009-1.pdf