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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.

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That is the generally accepted definition of an infinite summation. I do note, however, that sometimes we will not use that definition, as you can see... (I'd recommend using your definition, it is more common, also, the methods we use if the summation doesn't converge may be... less accepted as correct.) – Simple Art Feb 12 at 0:03
up vote 30 down vote accepted

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.

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Well, technically you must first define $\mathrm{lim}_{n \rightarrow \infty}$, which is most certainly not the traditional epsilon-delta definition. And then you have to make sure your definition is meaningful when we restrict $n$ to the natural numbers. But yeah, that's basically how infinite sums are defined. The rest can mostly be taken care of with Cauchy sequences. – Kevin Feb 12 at 4:29
@Kevin: What do you mean, "not the traditional epsilon-delta definition"? Why not? – TonyK Feb 12 at 10:45
@TonyK Probably because $|n - \infty| < \delta$ is meaningless. – kevinsa5 Feb 12 at 16:13
But $n\to\infty$ use a different traditional definition (most often $N$ and $\epsilon$). Saying it doesn't use $\delta$ is beside the point. – Teepeemm Feb 12 at 17:08

This is the definition of infinite series. It is the limit of the partial sums $S_n$:

$$S_n = \sum_{k = 0}^n a_k$$

$$\sum_{k = 0}^{\infty} a_k := \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k = 0}^n a_k$$

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