convergence tests for series:
- many are obvious:
terms must go to zero
comparison test
integral test (comparison with integral)
alternating series

ratio: small-scale comparison with geometric
root test: large-scale comparison with geometric
  ** these are the key ones, as they're used in power series
(for nice series, ratio test works...)

ratio: 
more properly: given a sequence $a_n$, the sum $\sum a_n$ converges if
$\limsup \abs{\frac{a_{n+1}}{a_n}}=L<1$
  (comparison with (a multiple of) the geometric series)
it diverges if
$\liminf \abs{\frac{a_{n+1}}{a_n}}=L>1$
  (obvious: terms don't go to zero!)
...and otherwise it's inconclusive.
(with $\lim$, it's called the ``root'' test; with $\limsup$ and
 $\liminf$, it's called the ``delicate root test'')


Here's a way to introduce it:
- we understand geometric series
- a constant times geometric series also works:
  and lo! it has constant ratio of successive terms
- even if the ratio isn't constant, but is (uniformly!) bounded away
  from 1 (say $\leq c < 1$), it's still less than -some- geometric. Which geometric?
  The bound: $c$
- actually, just need it to be -eventually- bounded ratio
decomposing, this says lim sup (ratio) < constant

NB: if ratio goes to $1$ (i.e., it's not less than some -fixed-
geometric), then dunno: consider $\zeta(1)$ and $\zeta(2)$

------------------------------------
root:
this is a -coarse- version of the ratio test:
the ratio test doesn't work if you have something like
1/2, 1/4, 1/8,
1/4, 1/8, 1/16,
1/8, 1/16, 1/32,...
(i.e., 2 steps forward, 1 step back)
b/c one step at a time, it doesn't work -- but overall it's okay,
and that's what the root test measures.

It's actually better than that:
suppose your terms are bounded by $f(x)/2^x$,
where $\sqrt[n]{f(n)} \to 1$ (i.e., $f$ grows subexponentially,
i.e., slower than any exponential); then the root test still works
  (and doesn't worry about lil' steps...)
  alternatively, this is because you can reduce the exponent in the
  base by an -arbitrarily small amount- and kill the numerator:
  $f(x)/2^x = (f(x)/(2/2-\epsilon)^x )/(2-\epsilon)^x$,
  and now the numerator goes to zero.

(also have ``delicate'' root test)
