Question

How do you test for the convergence or divergence of a non geometric series to infinity?

Answer 1

There are many different theorems providing tests and criteria to assess the convergence of a numeric series. Here are the most commonly used.

Given a series:

`sum_(n=0)^oo a_n`

the first important test is Cauchy's necessary condition stating that the series can converge only if `lim_(n->oo) a_n = 0`.

As this is a necessary condition, it can only prove that the series does not converge. We also have two important tests, based on the properties of `a_n` that can prove the series to converge or diverge:

Ratio test:

`lim_(n->oo) abs(a_n/a_(n+1)) < 1 <=> sum_(n=0)^oo a_n` is convergent

`lim_(n->oo) abs(a_n/a_(n+1)) > 1 <=> sum_(n=0)^oo a_n` is not convergent

if the limit is `1` the test is indecisive.

Root test

`lim_(n->oo) root(n)(a_n) < 1 <=> sum_(n=0)^oo a_n` is convergent

`lim_(n->oo) root(n)(a_n) > 1 <=> sum_(n=0)^oo a_n` is not convergent

if the limit is `1` the test is indecisive.

Moreover, we can often proceed by comparing the series with some other series that we now to be convergent or divergent.

Squeeze theorem

If `a_n <= b_n <= c_n` and both `sum_(n=0)^oo a_n` and `sum_(n=0)^oo c_n` are convergent, then also `sum_(n=0)^oo b_n` is convergent.

If the series has positive terms (or, which is equivalent, when we test for absolute convergence), this can be generalized as the:

Direct comparison test

if `a_n, b_n >=0` with `a_n <= b_n` and `L` is finite, then

then:

`sum_(n=0)^oo b_n = L => sum_(n=0)^oo a_n` is convergent

`sum_(n=0)^oo a_n = oo => sum_(n=0)^oo b_n = oo`

To this purpose, beside the geometric series we have an important test series provided by the `p`-series theorem, stating that:

p-series Test

`sum_(n=0)^oo 1/n^p`

is convergent for `p>1` and divergent for `p<=1`

The comparison test is further generalized by the:

Limit comparison test

Given two series `sum_(n=0)^oo a_n`, `sum_(n=0)^oo b_n`

if `lim_(n->oo) a_n/b_n = L` with `0 < L < oo` then both series have the same character, that is either both are convergent or both divergent;

if `lim_(n->oo) a_n/b_n = 0` then if `sum_(n=0)^oo b_n` is convergent also `sum_(n=0)^oo a_n` is convergent;

if `lim_(n->oo) a_n/b_n = oo` then if `sum_(n=0)^oo b_n` is divergent also `sum_(n=0)^oo a_n` is divergent;

Finally, a special case is the one of alternating series in the form:

`sum_(n=0)^oo (-1)^na_n`

with `a_n >=0`. In such case we can use:

Leibniz' theorem

The series:

`sum_(n=0)^oo (-1)^na_n`

is convergent if:

`(1) " " lim_(n->oo) a_n = 0`

`(2)" " a_n > a_(n+1)` for `n > N`