In Big-O notation, do we take the limit of a series's terms as we approach infinity, or evaluate the term at infinity?

Question

For example, say we had a function:

$$f(x)=x+2x^2+3x^3+4x^4+...\text{ for } x\geq1$$

Then can we write:

$$f(x)=O(x) \text{ as } x\rightarrow1$$

since the value of $f(x)$ at $x=1$ is $-1/12$? Or do we argue that the limit of the function is $\infty$, thererefore there is no bounding term?

Answer 1

Regarding your mathematical question about the big "O" notation:

$$f(x)=\mathcal O(x)\;\;\text{as}\;\;x\to1\iff \lim\sup_{x\to1}\left|\frac{f(x)}x\right|<\infty\iff$$

$$\iff\lim\sup_{x\to1}(1+2x+3x^2+\ldots)<\infty\;,\;\;\text{and this is clearly false}.$$

under the usual assumptions and definition in mathematics.

Answer 2

To evaluate the function, multiply it by $(1-x)$ to get

$$\begin{align} (1-x)f(x)=&(1-x)(x+2x^2+3x^3+\dots)\\ (1)\qquad=&1(x+2x^2+3x^3+\dots)-x(x+2x^2+3x^3+\dots)\\ (2)\qquad=&(x+2x^2+3x^3+\dots)-(x^2+2x^3+3x^4+\dots)\\ (3)\qquad=&x+(2-1)x^2+(3-2)x^3+\dots\\ (4)\qquad=&x+x^2+x^3+\dots\\ (5)\qquad=&\frac x{1-x} \end{align}$$

$(1):$ distribute the $(1-x)$.

$(2):$ distribute the $x$ on the second series.

$(3):$ combine like terms.

$(4):$ calculate like terms.

$(5):$ evaluate geometric series.

$$(1-x)f(x)=\frac x{1-x}$$

$$f(x)=\frac x{(1-x)^2}$$

So even if you could have the function defined for $|x|\ge1$, $f(1)$ still wouldn't exist and we wouldn't have the silly $f(1)=-1/12$ thing.

Now, there are values you can pull out the function for $x=1$, but they do not relate to $-1/12$ nor do they have any meaning here. For example, the residue of the function at $x=1$ is given by $1$ (but who cares?)