Question

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

Answer 1

An infinite geometric series is the sum of an infinite geometric sequence. This series would have no last term. The general form of the infinite geometric series is $a + ar + a{r^2} + a{r^3} + ...$ where $a$ is the first term and $r$ is the common ratio.
There are many different theorems providing tests and criteria to assess the convergence of a numeric series.

Complete step by step answer:
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:
$\sum\limits_{n = 0}^\infty {{a_n}}$
The first important test is Cauchy's necessary condition stating that the series can converge only if $\mathop {\lim }\limits_{n \to 0} {a_n} = 0$
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: $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_n}}}{{{a_{n + 1}}}}} \right| < 1 \Leftrightarrow \sum\limits_{n = 0}^\infty {{a_n}}$ is convergent.
$\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_n}}}{{{a_{n + 1}}}}} \right| > 1 \Leftrightarrow \sum\limits_{n = 0}^\infty {{a_n}}$is divergent.
If the limit is 1 the test is indecisive
Root test: $\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} < 1 \Leftrightarrow \sum\limits_{n = 0}^\infty {{a_n}}$is convergent.
$\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} > 1 \Leftrightarrow \sum\limits_{n = 0}^\infty {{a_n}}$is divergent.
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.

Note: There are few more processes to show.
Squeeze theorem: If ${a_n} < {b_n} < {c_n}$and both $\sum\limits_{n = 0}^\infty {{a_n}}$and $\sum\limits_{n = 0}^\infty {{c_n}}$are convergent, then also $\sum\limits_{n = 0}^\infty {{b_n}}$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 ${a_n},{b_n} \geqslant 0$with ${a_n} \leqslant {b_n}$and L is finite, then
then:$\sum\limits_{n = 0}^\infty {{b_n}} = L = \sum\limits_{n = 0}^\infty {{a_n}}$ is convergent.
$\sum\limits_{n = 0}^\infty {{a_n}} = \infty \Rightarrow \sum\limits_{n = 0}^\infty {{b_n}} = \infty$
To this purpose, beside the geometric series we have an important test series provided by the p-series theorem, stating that:
p-series Test
$\sum\limits_{n = 0}^\infty {\dfrac{1}{{{n^p}}}}$is convergent for p>1 and divergent for p≤1.