I've found my textbook resorting to things like the ratio test with, for example, the series $$\sum_{n=0}^{\infty} x^{n}/n!$$
Could that series not be considered a geometric series, where $a = 1/n! $ and $r=x$ ? Is it not possible to have $n$ be a part of $a$ (i.e. does $a$ always have to be some constant real number)?
Or maybe the problem is that $n$ began at $0$, or even the power on $x$? In that case, would $$\sum_{n=1}^{\infty} x^{n-1}/n!$$ be a geometric series? If not, why?
Any help is appreciated!
To have a geometric series $\sum ar^n$ you have a fixed "$a$" and fixed ratio "$r$" so that the ratio of any two successive terms is: $(ar^{k+1})/(ar^k) = r$ (a fixed amount). This ratio is independent of our choice of index $k$.
So by fixed I mean fixed relative to the index variable.
You can't let $a=1/n!$ since $a$ changes as the index variable $n$ changes.
Notice that the ratio of two successive terms is $(x^{k+1}/(k+1)!)/(x^k/k!) = x/k$. This ratio changes as $k$ changes.
Thus this is not a geometric series.
The issue isn't whether $a$ and $r$ are real numbers, complex numbers, variables, matrices or any other kind of object. The issue lies with whether the ratio of two successive terms is the same regardless of which successive terms we consider.