TITLE: Summation of a term to infinity
QUESTION [15 upvotes]: I read through many tutorials but no one mentioned this explicitly. 
Is the following conversion valid?
$$\sum_{k=0}^\infty \frac{k-1}{2^k} = \lim_{n\to \infty} \sum_{k=0}^n \frac{k-1}{2^k}$$
Please excuse if it seems stupid or too simple to ask in the forum.

REPLY [29 votes]: It's not only valid, it's how it's defined.

Note that the operation "addition" is defined only if we apply it a finite amount of times. Thus, adding an infinite amount of terms doesn't make sense. We'll have to define it as a limit, as that only includes nice, finite sums.

REPLY [16 votes]: This is the definition of infinite series. It is the limit of the partial sums $S_n$:
$$S_n = \sum_{k = 0}^n a_k$$
$$\sum_{k = 0}^{\infty} a_k := \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k = 0}^n a_k$$