Is plugging numbers into the ratio test allowed? So I'm going through the notes for my online summer calculus class, and something struck me as odd about the ratio test: it used variables only, there were no numbers plugged in. For example, in the series $$\sum_{n=1}^\infty\frac{n}{4^n}$$ we have $a_{n+1}=\frac{n+1}{4^{n+1}}$ and $a_n=\frac{n}{4^n}$ as the numerator and denominator in the ratio $ \left | \frac{a_{n+1}}{a_n} \right | $.
Now, my notes say that we then plug $a_{n+1}=\frac{n+1}{4^{n+1}}$ and $a_{n}=\frac{n}{4^n}$ directly into the ratio, giving us
$$ \left | \frac{\frac{n+1}{4^{n+1}}}{\frac{n}{4^n}} \right | $$
which we simplify down to $$ \left | \frac{n+1}{4n} \right | $$
That ratio is then plugged into the limit and which we find is $\frac{1}{4}$, so we know that the series converges since $\frac{1}{4} < 1$.
My question is, why do we solve for the initial ratio algebraically with symbols? Couldn't we just pick an integer $n$ and plug that as well as the integer $n+1$ into the equation to get our ratio? It seems like a lot of extra work that can be simplified.

No, you can't just plug in numbers. 
The ratio test is supposed to be about $\lim \dfrac{|a_{n+1}|}{|a_{n}|}$, and this limit is the key part. 
EDIT
Let's look into this a bit more. Suppose I have the series whose first thousand terms $a_1, \ldots, a_{1000}$ are $1,2, \ldots, 1000$, i.e. $a_n = n$. But after those terms, the series becomes $a_n = \dfrac{1}{2^n + n}$. If you were to just 'find the ratio' of any two of the first 1000 terms, you would think the series diverges, since you'd get a number that's greater than $1$. And if you were to find the ratio of any two of the subsequent terms, you'd get different ratios for every pair! But the limiting ratio is nonetheless $1/2$, which explains why $\sum a_n$ converges in the end.
