Q:

Ratio test, Root test, and Divergence test related.

(I) Ratio test: If the result is smaller than 1 then the sum is convergent, and if the sum is larger than 1 then the sum is divergent, and that got me thinking if negative infinity (smaller than 1) and positive infinity (larger than 1) are convergent and divergent, respectively. (And I understand that infinity is not a number, but still isn't negative infinity smaller than one? Or have I gone bananas?)
(II) Root test: If the result is smaller than 1 then the sum is (absolutely) convergent, but if the result is larger than 1 or equal to infinity then the sum is divergent. What kind of infinity do they mean in this case (Root test)?
(III) Divergence test: If the limit is not equal to zero, then the sum will diverge (and I understand that if it does equal zero then it can still be divergent, but anyway). Now, what I want to know is does this include infinity, be it negative or positive infinity, or am I missing something here?
Short note, most of the exercises I am required to do require detailed explanations and even if you do the whole calculations correctly but your explanation is wrong then the whole thing is considered wrong. Therefore, I want to actually know what I am doing, thank you! Also, please use (I)-(III) when explaining a specific test so I know which you're talking about.

A:

(I) You need the absolute value of the limit result to be less than 1, so if your limit is $\pm\infty$ you will get a divergent series.
(II) You can think of any value greater than 1 as good enough to prove divergence.  Any "kind of infinity" will do.  Once again, absolute value applies here.

The takeaway from ratio and root tests is threefold:

$|\text{Limit}|<1\Longrightarrow\text{series converges absolutely}$
$|\text{Limit}|>1\Longrightarrow\text{series diverges}$
$|\text{Limit}|=1\Longrightarrow\text{test is inconclusive, so use another test for convergence!}$

(III) If the limit of the general term is not zero, the series diverges.  If the limit is zero, the test is inconclusive! Be careful that you do not use the converse of this statement, because the converse is not true.
Examples:

$\displaystyle\sum_{n=1}^\infty\frac{n}{n+1}$ diverges since $\displaystyle\lim_{n\to\infty}\frac{n}{n+1}=1\ne0.$
$\displaystyle\sum_{n=1}^\infty\frac{n^2}{n+1}$ diverges since
$\displaystyle\lim_{n\to\infty}\frac{n}{n+1}=\infty\ne0.$
$\displaystyle\sum_{n=1}^\infty\frac{1}{n+1}$ has $\displaystyle\lim_{n\to\infty}\frac{1}{n+1}=0$ which means we do not know if the series converges or diverges unless we use another test.

