Question

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

Answer 1

Hint: 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.