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

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#