Editing Ratio test (section)

== Proof ==
[[File:Ratio test proof.svg|thumb|300px|In this example, the ratio of adjacent terms in the blue sequence converges to L=1/2. We choose ''r''&nbsp;= (L+1)/2&nbsp;= 3/4. Then the blue sequence is dominated by the red sequence {{mvar|r<sup>k</sup>}} for all ''n'' ≥ 2. The red sequence converges, so the blue sequence does as well.]]
Below is a proof of the validity of the generalized ratio test.

Suppose that <math>r=\liminf_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|>1</math>. We also suppose that <math>(a_n)</math> has infinite non-zero members, otherwise the series is just a finite sum hence it converges. Then there exists some <math>\ell\in(1;r)</math> such that there exists a natural number <math>n_0\ge2</math> satisfying <math>a_{n_0}\ne0</math> and <math>\left|\frac{a_{n+1}}{a_n}\right|>\ell</math> for all <math>n\ge n_0</math>, because if no such <math>\ell</math> exists then there exists arbitrarily large <math>n</math> satisfying <math>\left|\frac{a_{n+1}}{a_n}\right|<\ell</math> for every <math>\ell\in(1;r)</math>, then we can find a subsequence <math>\left(a_{n_k}\right)_{k=1}^\infty</math> satisfying <math>\limsup_{n\to\infty}\left|\frac{a_{n_k+1}}{a_{n_k}} \right|\le\ell<r</math>, but this contradicts the fact that <math>r</math> is the [[limit inferior]] of <math>\left|\frac{a_{n+1}}{a_n}\right|</math> as <math>n\to\infty</math>, implying the existence of <math>\ell</math>. Then we notice that for <math>n\ge n_0+1</math>, <math>|a_n|> \ell|a_{n-1}|>\ell^2|a_{n-2}|>...>\ell^{n-n_0}\left|a_{n_0}\right|</math>. Notice that <math>\ell>1</math> so <math>\ell^n\to\infty</math> as <math>n\to\infty</math> and <math>\left|a_{n_0}\right|>0</math>, this implies <math>(a_n)</math> diverges so the series <math>\sum_{n=1}^\infty a_n</math> diverges by the [[n-th term test]].<br>
Now suppose <math>R=\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1</math>. Similar to the above case, we may find a natural number <math>n_1</math> and a <math>c\in(R;1)</math> such that <math>|a_n|\le c^{n-n_1}\left|a_{n_1}\right|</math> for <math>n\ge n_1</math>. Then
<math display="block">\sum_{n=1}^\infty |a_n|=\sum_{k=1}^{n_1-1}|a_k|+\sum_{n=n_1}^\infty |a_n|\le\sum_{k=1}^{n_1-1}|a_k|+\sum_{n=n_1}^\infty c^{n-n_1}|a_{n_1}|=\sum_{k=1}^{n_1-1}|a_k|+\left|a_{n_1}\right|\sum_{n=0}^\infty c^n.</math>
The series <math>\sum_{n=0}^\infty c^n</math> is the [[geometric series]] with common ratio <math>c\in(0;1)</math>, hence <math>\sum_{n=0}^\infty c^n=\frac{1}{1-c}</math> which is finite. The sum <math>\sum_{k=1}^{n_1-1}|a_k|</math> is a finite sum and hence it is bounded, this implies the series <math>\sum_{n=1}^\infty |a_n|</math> converges by the [[monotone convergence theorem]] and the series <math>\sum_{n=1}^\infty a_n</math> converges by the absolute convergence test.<br>
When the limit <math>\left|\frac{a_{n+1}}{a_n}\right|</math> exists and equals to <math>L</math> then <math>r=R=L</math>, this gives the original ratio test.