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{{Calculus|Series}}

In [[mathematics]], the '''ratio test''' is a [[convergence tests|test]] (or "criterion") for the [[convergent series|convergence]] of a [[series (mathematics)|series]]

:<math>\sum_{n=1}^\infty a_n,</math>

where each term is a real or complex number and {{mvar|a<sub>n</sub>}} is nonzero when {{mvar|n}} is large. The test was first published by [[Jean le Rond d'Alembert]] and is sometimes known as '''d'Alembert's ratio test '''or as the '''Cauchy ratio test'''.<ref name=":0">{{mathworld|title=Ratio Test|urlname=RatioTest}}</ref>

== The test ==
[[File:Decision diagram for the ratio test.svg|thumb|Decision diagram for the ratio test]]
The usual form of the test makes use of the [[limit (mathematics)|limit]]
{{NumBlk|:|<math>L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.</math>|{{EquationRef|1}}}}
The ratio test states that:
* if ''L'' < 1 then the series [[absolute convergence|converges absolutely]]; 
* if ''L'' > 1 then the [[divergent series|series is divergent]];
* if ''L'' = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

It is possible to make the ratio test applicable to certain cases where the limit ''L'' fails to exist, if [[limit superior]] and [[limit inferior]] are used. The test criteria can also be refined so that the test is sometimes conclusive even when ''L'' = 1. More specifically, let

:<math>R = \lim\sup \left|\frac{a_{n+1}}{a_n}\right|</math>
:<math>r = \lim\inf \left|\frac{a_{n+1}}{a_n}\right|</math>.

Then the ratio test states that:<ref>{{harvnb|Rudin|1976|loc=§3.34}}</ref><ref>{{harvnb|Apostol|1974|loc=§8.14}}</ref>
* if ''R'' < 1, the series converges absolutely;
* if ''r'' > 1, the series diverges;
* if <math>\left|\frac{a_{n+1}}{a_n}\right|\ge 1</math> for all large ''n'' (regardless of the value of ''r''), the series also diverges; this is because <math>|a_n|</math> is nonzero and increasing and hence {{mvar|a<sub>n</sub>}} does not approach zero;
* the test is otherwise inconclusive.

If the limit ''L'' in ({{EquationNote|1}}) exists, we must have ''L'' = ''R'' = ''r''. So the original ratio test is a weaker version of the refined one.

== Examples ==
=== Convergent because ''L'' < 1 ===
Consider the series

:<math>\sum_{n=1}^\infty\frac{n}{e^n}</math>

Applying the ratio test, one computes the limit

:<math>L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right| = \frac{1}{e} < 1.</math>

Since this limit is less than 1, the series converges.

=== Divergent because ''L'' > 1 ===
Consider the series

:<math>\sum_{n=1}^\infty\frac{e^n}{n}.</math>

Putting this into the ratio test:

:<math>L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}} \right|
= e > 1.</math>

Thus the series diverges.

=== Inconclusive because ''L'' = 1 ===
Consider the three series

:<math>\sum_{n=1}^\infty 1,</math>
:<math>\sum_{n=1}^\infty \frac{1}{n^2},</math>
:<math>\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}.</math>

The first series ([[1 + 1 + 1 + 1 + ⋯]]) diverges, the second one (the one central to the [[Basel problem]]) converges absolutely and the third one (the [[alternating harmonic series]]) converges conditionally. However, the term-by-term magnitude ratios <math>\left|\frac{a_{n+1}}{a_n}\right|</math> of the three series are respectively <math>1,</math> &nbsp;&nbsp;<math>\frac{n^2}{(n+1)^2}</math> &nbsp;&nbsp; and  &nbsp;&nbsp;<math>\frac{n}{n+1}</math>. So, in all three cases, one has that the limit <math>\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|</math> is equal to 1. This illustrates that when ''L'' = 1, the series may converge or diverge, and hence the original ratio test is inconclusive.  In such cases, more refined tests are required to determine convergence or divergence.

== 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 original ratio test.

Suppose that <math>L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_{n}}\right| < 1</math>. We can then show that the series converges absolutely by showing that its terms will eventually become less than those of a certain convergent [[geometric series]]. To do this, let <math>r = \frac{L+1}{2}</math>.  Then ''r'' is strictly between ''L'' and 1, and <math>|a_{n+1}| < r |a_{n}|</math> for sufficiently large ''n''; say, for all ''n'' greater than ''N''. Hence <math>|a_{n+i}| < r^i|a_{n}|</math> for each ''n'' > ''N'' and ''i'' > 0, and so

:<math>\sum_{i=N+1}^{\infty}|a_i| = \sum_{i=1}^{\infty} \left |a_{N+i} \right | < \sum_{i=1}^{\infty}r^{i}|a_{N}| = |a_{N}| \sum_{i=1}^{\infty} r^{i} = |a_{N}|\frac{r}{1 - r} < \infty.</math>

That is, the series converges absolutely.

On the other hand, if ''L'' > 1, then <math>|a_{n+1}| > |a_{n}|</math> for sufficiently large ''n'', so that the limit of the summands is non-zero. Hence the series diverges.

== Extensions for ''L'' = 1 ==

As seen in the previous example, the ratio test may be inconclusive when the limit of the ratio is 1. Extensions to the ratio test, however, sometimes allows one to deal with this case.<ref name="Bromwich1908">{{cite book |last=Bromwich |first=T. J. I’A |date=1908 |title=An Introduction To The Theory of Infinite Series |url= |location= |publisher=Merchant Books |page= |isbn= |author-link= Thomas John I'Anson Bromwich}}</ref><ref name="Knopp">{{cite book |last=Knopp |first=Konrad |date=1954 |title=Theory and Application of Infinite Series |url=https://archive.org/details/theoryandapplica031692mbp/page/n5 |location=London |publisher=Blackie & Son Ltd. |page= |isbn= |author-link=Konrad Knopp }}</ref><ref name="Tong1994">
{{cite journal |last1=Tong |first1=Jingcheng|date=May 1994|title=Kummer's Test Gives Characterizations for Convergence or Divergence of all Positive Series|jstor=2974907|journal=The American Mathematical Monthly |volume=101 |issue=5 |pages=450–452 |doi=10.2307/2974907 }}</ref><ref name="Ali2008">
{{cite journal |last1=Ali |first1=Sayel A.|date=2008 |title=The mth Ratio Test: New Convergence Test for Series |url= http://web.mnstate.edu/alis/mth_ratio_test_monthly514-524.pdf|journal=The American Mathematical Monthly |volume=115 |issue=6 |pages=514–524 |doi= 10.1080/00029890.2008.11920558|access-date=21 November 2018 }}</ref><ref name="Samelson1995">
{{cite journal |last1=Samelson |first1=Hans|date=November 1995|title=More on Kummer's Test|jstor=2974510|journal=The American Mathematical Monthly |volume=102 |issue=9 |pages=817–818 |doi=10.2307/2974510 }}</ref><ref name="Blackburn2012">{{cite web |url= http://sites.math.washington.edu/~morrow/336_12/papers/kyle.pdf|title= The mth Ratio Convergence Test and Other Unconventional Convergence Tests|last= Blackburn|first=Kyle |date=4 May 2012 |website= |publisher=University of Washington College of Arts and Sciences |access-date= 27 November 2018 |quote=}}</ref><ref name="Duris2009">{{cite thesis |last1=Ďuriš |first1=František |date=2009 |title=Infinite series: Convergence tests |type=Bachelor's thesis |publisher=Katedra Informatiky, Fakulta Matematiky, Fyziky a Informatiky, Univerzita Komenského, Bratislava |url= http://www.dcs.fmph.uniba.sk/bakalarky/obhajene/Detail.php?id=90|access-date= 28 November 2018 }}</ref><ref name="Duris2018">{{cite arxiv |last1=Ďuriš |first1=František |date= 2 February 2018 |title=On Kummer's test of convergence and its relation to basic comparison tests |eprint=1612.05167 |class=math.HO }}</ref>

In all the tests below one assumes that Σ''a''<sub>''n''</sub> is a sum with positive ''a''<sub>''n''</sub>. These tests also may be applied to any series with a finite number of negative terms. Any such series may be written as:

:<math>\sum_{n=1}^\infty a_n = \sum_{n=1}^N a_n+\sum_{n=N+1}^\infty a_n</math>

where ''a<sub>N</sub>'' is the highest-indexed negative term. The first expression on the right is a partial sum which will be finite, and so the convergence of the entire series will be determined by the convergence properties of the second expression on the right, which may be re-indexed to form a series of all positive terms beginning at ''n''=1.

Each test defines a test parameter (ρ<sub>n</sub>) which specifies the behavior of that parameter needed to establish convergence or divergence. For each test, a weaker form of the test exists which will instead place restrictions upon lim<sub>n->∞</sub>ρ<sub>n</sub>.

All of the tests have regions in which they fail to describe the convergence properties of ∑a<sub>n</sub>. In fact, no convergence test can fully describe the convergence properties of the series.<ref name="Bromwich1908"/><ref name="Duris2009"/> This is because if ∑a<sub>n</sub> is convergent, a second convergent series ∑b<sub>n</sub> can be found which converges more slowly: i.e., it has the property that lim<sub>n->∞</sub> (b<sub>n</sub>/a<sub>n</sub>) = ∞. Furthermore, if ∑a<sub>n</sub> is divergent, a second divergent series ∑b<sub>n</sub> can be found which diverges more slowly: i.e., it has the property that lim<sub>n->∞</sub> (b<sub>n</sub>/a<sub>n</sub>) = 0. Convergence tests essentially use the comparison test on some particular family of a<sub>n</sub>, and fail for sequences which converge or diverge more slowly.

=== De Morgan hierarchy ===

[[Augustus De Morgan]] proposed a hierarchy of ratio-type tests<ref name="Bromwich1908"/><ref name="Blackburn2012"/>

The ratio test parameters (<math>\rho_n</math>) below all generally involve terms of the form <math>D_n a_n/a_{n+1}-D_{n+1}</math>. This term may be multiplied by <math>a_{n+1}/a_n</math> to yield <math>D_n-D_{n+1}a_{n+1}/a_n</math>. This term can replace the former term in the definition of the test parameters and the conclusions drawn will remain the same. Accordingly, there will be no distinction drawn between references which use one or the other form of the test parameter.

====1. d’Alembert’s ratio test====

The first test in the De Morgan hierarchy is the ratio test as described above.

====2. Raabe's test====

This extension is due to [[Joseph Ludwig Raabe]]. Define:

:<math>\rho_n \equiv n\left(\frac{a_n}{a_{n+1}}-1\right)</math>

(and some extra terms, see Ali, Blackburn, Feld, Duris (none), Duris2)

The series will:<ref name="Ali2008"/><ref name="Duris2009"/><ref name="Blackburn2012"/>
* Converge when there exists a ''c>1'' such that <math>\rho_n \ge c</math> for all ''n>N''.
* Diverge when <math>\rho_n \le 1</math> for all ''n>N''.
* Otherwise, the test is inconclusive.

For the limit version,<ref>{{mathworld|title=Raabe's Test|urlname=RaabesTest}}</ref> the series will:
* Converge if <math>\lim_{n\to\infty}\rho_n>1</math> (this includes the case ''ρ'' = ∞)
* Diverge if <math>\lim_{n\to\infty}\rho_n<1</math>.
* If ''ρ'' = 1, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior.<ref  name="Bromwich1908"/> The series will:
* Converge if <math>\liminf_{n \to \infty} \rho_n > 1</math>
* Diverge if <math>\limsup_{n \rightarrow \infty} \rho_n < 1</math>
* Otherwise, the test is inconclusive.

=====Proof of  Raabe's test=====

Defining <math>\rho_n \equiv n\left(\frac{a_n}{a_{n+1}}-1\right)</math>, we need not assume the limit exists; if <math>\limsup\rho_n<1</math>, then <math>\sum a_n</math> diverges, while if <math>\liminf \rho_n>1</math> the sum converges.

The proof proceeds essentially by comparison with <math>\sum1/n^R</math>. Suppose first that <math>\limsup\rho_n<1</math>. Of course
if <math>\limsup\rho_n<0</math> then <math>a_{n+1}\ge a_n</math> for large <math>n</math>, so the sum diverges; assume then that <math>0\le\limsup\rho_n<1</math>. There exists <math>R<1</math> such that <math>\rho_n\le R</math> for all <math>n\ge N</math>, which is to say that <math>a_{n}/a_{n+1}\le \left(1+\frac Rn\right)\le e^{R/n}</math>. Thus <math>a_{n+1}\ge a_ne^{-R/n}</math>, which implies that
<math>a_{n+1}\ge a_Ne^{-R(1/N+\dots+1/n)}\ge ca_Ne^{-R\log(n)}=ca_N/n^R</math> for <math>n\ge N</math>; since <math>R<1</math> this shows that <math>\sum a_n</math> diverges.

The proof of  the other half is entirely analogous, with most of the inequalities simply reversed. We need a preliminary inequality to use
in place of the simple <math>1+t<e^t</math> that was used above:  Fix <math>R</math> and <math>N</math>. Note that
<math>\log\left(1+\frac Rn\right)=\frac Rn+O\left(\frac 1{n^2}\right)</math>. So <math>\log\left(\left(1+\frac RN\right)\dots\left(1+\frac Rn\right)\right)
=R\left(\frac 1N+\dots+\frac 1n\right)+O(1)=R\log(n)+O(1)</math>; hence <math>\left(1+\frac RN\right)\dots\left(1+\frac Rn\right)\ge cn^R</math>.

Suppose now that <math>\liminf\rho_n>1</math>. Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there exists <math>R>1</math> such that <math>a_{n+1}\le ca_Nn^{-R}</math> for <math>n\ge N</math>; since <math>R>1</math> this shows that <math>\sum a_n</math> converges.

==== 3. Bertrand’s test ====

This extension is due to [[Joseph Bertrand]] and [[Augustus De Morgan]].

Defining:

:<math>\rho_n \equiv n \ln n\left(\frac{a_n}{a_{n+1}}-1\right)-\ln n</math>

Bertrand's test<ref name="Bromwich1908"/><ref name="Duris2009"/> asserts that the series will:
* Converge when there exists a ''c>1'' such that <math>\rho_n \ge c</math> for all ''n>N''.
* Diverge when <math>\rho_n \le 1</math> for all ''n>N''.
* Otherwise, the test is inconclusive.

For the limit version, the series will:
* Converge if <math>\lim_{n\to\infty}\rho_n>1</math> (this includes the case ''ρ'' = ∞)
* Diverge if <math>\lim_{n\to\infty}\rho_n<1</math>.
* If ''ρ'' = 1, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior.<ref name="Bromwich1908"/><ref name="Blackburn2012"/><ref>{{mathworld|title=Bertrand's Test|urlname=BertrandsTest}}</ref> The series will:
* Converge if <math>\liminf \rho_n > 1</math>
* Diverge if  <math>\limsup \rho_n < 1</math>
* Otherwise, the test is inconclusive.

==== 4. Gauss’s test ====

This extension is due to [[Carl Friedrich Gauss]].

Assuming ''a<sub>n</sub>'' > 0 and ''r > 1'', if a bounded sequence ''C<sub>n</sub>'' can be found such that for all ''n'':<ref name="Knopp"/><ref name="Ali2008"/><ref name="Blackburn2012"/><ref name="Duris2009"/>

:<math>\frac{a_n}{a_{n+1}}=1+\frac{\rho}{n}+\frac{C_n}{n^r}</math>

then the series will:
* Converge if <math>\rho>1</math>
* Diverge if <math>\rho \le 1</math>

==== 5. Kummer’s test ====

This extension is due to [[Ernst Kummer]].

Let ζ<sub>''n''</sub> be an auxiliary sequence of positive constants. Define

:<math>\rho_n \equiv \left(\zeta_n \frac{a_n}{a_{n+1}} - \zeta_{n+1}\right)</math>

Kummer's test states that the series will:<ref name="Knopp"/><ref name="Tong1994"/><ref name="Duris2009"/><ref name="Duris2018"/>
* Converge if there exists a <math>c>0</math> such that <math>\rho_n \ge c</math> for all n>N. (Note this is not the same as saying <math>\rho_n > 0</math>)
* Diverge if <math>\rho_n \le 0</math> for all n>N and <math>\sum_{n=1}^\infty 1/\zeta_n</math> diverges.

For the limit version, the series will:<ref>{{mathworld|title=Kummer's Test|urlname=KummersTest}}</ref><ref name="Ali2008"/><ref name="Blackburn2012"/>
* Converge if <math>\lim_{n\to\infty}\rho_n>0</math> (this includes the case ''ρ'' = ∞)
* Diverge if <math>\lim_{n\to\infty}\rho_n<0</math> and <math>\sum_{n=1}^\infty 1/\zeta_n</math> diverges.
* Otherwise the test is inconclusive

When the above limit does not exist, it may be possible to use limits superior and inferior.<ref name="Bromwich1908"/> The series will
* Converge if <math>\liminf_{n \to \infty} \rho_n >0</math>
* Diverge if <math>\limsup_{n \to \infty} \rho_n <0</math> and <math>\sum 1/\zeta_n</math> diverges.

===== Special cases =====

All of the tests in De Morgan's hierarchy except Gauss's test can easily be seen as special cases of Kummer's test:<ref name="Bromwich1908"/>
* For the ratio test, let ζ<sub>n</sub>=1. Then:
::<math>\rho_{Kummer} = \left(\frac{a_n}{a_{n+1}}-1\right) = 1/\rho_{Ratio}-1</math>
* For Raabe's test, let ζ<sub>n</sub>=n. Then:
::<math>\rho_{Kummer} = \left(n\frac{a_n}{a_{n+1}}-(n+1)\right) = \rho_{Raabe}-1</math>
* For Bertrand's test, let ζ<sub>n</sub>=n ln(n). Then:
::<math>\rho_{Kummer} =  n \ln(n)\left(\frac{a_n}{a_{n+1}}-1\right)-(n+1)\ln(n+1)</math>
:Using <math>\ln(n+1)=\ln(n)+\ln(1+1/n)</math> and approximating <math>\ln(1+1/n)\rightarrow 1/n</math> for large ''n'', which is negligible compared to the other terms, ρ<sub>Kummer</sub> may be written:
::<math>\rho_{Kummer} =  n \ln(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\ln(n)-1 = \rho_{Bertrand}-1</math>

Note that for these three tests, the higher they are in the De Morgan hierarchy, the more slowly the 1/ζ<sub>n</sub> series diverges.

=====Proof of Kummer's test=====

If <math>\rho_n>0</math> then fix a positive number <math>0<\delta<\rho_n</math>.  There exists
a natural number <math>N</math> such that for every <math>n>N,</math>
:<math>\delta\leq\zeta_{n}\frac{a_{n}}{a_{n+1}}-\zeta_{n+1}.</math>
Since <math>a_{n+1}>0</math>, for every <math>n> N,</math>
:<math>0\leq \delta a_{n+1}\leq \zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}.</math>
In particular <math>\zeta_{n+1}a_{n+1}\leq \zeta_{n}a_{n}</math> for all <math>n\geq N</math> which means that starting from the index
<math>N</math>
the sequence <math>\zeta_{n}a_{n}>0</math> is monotonically decreasing and
positive which in particular implies that it is bounded below by 0. Therefore, the limit
:<math>\lim_{n\to\infty}\zeta_{n}a_{n}=L</math> exists.
This implies that the positive  [[telescoping series]]
:<math>\sum_{n=1}^{\infty}\left(\zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}\right)</math> is convergent,
and since for all <math>n>N,</math>
:<math>\delta a_{n+1}\leq \zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}</math>
by the [[direct comparison test]] for positive series, the series
<math>\sum_{n=1}^{\infty}\delta a_{n+1}</math> is convergent.

On the other hand, if <math>\rho<0</math>, then there is an ''N'' such that <math>\zeta_n a_n</math> is increasing for <math>n>N</math>.  In particular, there exists an <math>\epsilon>0</math> for which <math>\zeta_n a_n>\epsilon</math> for all <math>n>N</math>, and so <math>\sum_n a_n=\sum_n \frac{a_n\zeta_n}{\zeta_n}</math> diverges by comparison with <math>\sum_n \frac \epsilon {\zeta_n}</math>.

=== Second ratio test ===
A more refined ratio test is the second ratio test:<ref name="Ali2008"/><ref name="Blackburn2012"/>
For <math>a_n>0</math> define:
{|
|<math>L_0\equiv\lim_{n\rightarrow \infty} \frac{a_{2n}}{a_n}</math>
|-
|<math>L_1\equiv\lim_{n\rightarrow \infty} \frac{a_{2n+1}}{a_n}</math>
|-
|<math>L\equiv\max(L_0,L_1)</math>
|}

By the second ratio test, the series will:
* Converge if <math>L<\frac{1}{2}</math>
* Diverge if <math>L>\frac{1}{2}</math>
* If <math>L=\frac{1}{2}</math> then the test is inconclusive.

If the above limits do not exist, it may be possible to use the limits superior and inferior. Define:

{|
|<math>L_0\equiv\limsup_{n\rightarrow \infty} \frac{a_{2n}}{a_n}</math>
|
|<math>L_1\equiv\limsup_{n\rightarrow \infty} \frac{a_{2n+1}}{a_n}</math>
|-
|<math>\ell_0\equiv\liminf_{n\rightarrow \infty} \frac{a_{2n}}{a_n}</math>
|
|<math>\ell_1\equiv\liminf_{n\rightarrow \infty} \frac{a_{2n+1}}{a_n}</math>
|-
|<math>L\equiv\max(L_0,L_1)</math>
|
|<math>\ell\equiv\min(\ell_0,\ell_1)</math>
|}

Then the series will:
* Converge if <math>L<\frac{1}{2}</math>
* Diverge if <math>\ell>\frac{1}{2}</math>
* If <math>\ell \le \frac{1}{2} \le L</math> then the test is inconclusive.

The second ratio test can be generalized to an ''m''-th ratio test, but higher orders are not found to be as useful.<ref name="Ali2008"/><ref name="Blackburn2012"/>

==See also==
* [[Root test]]
* [[Radius of convergence]]

== Footnotes ==
{{reflist}}

== References ==
*{{citation | last1=d'Alembert| first1=J.|authorlink=Jean le Rond d'Alembert| year=1768 | title=Opuscules |volume=V|pages=171–183| url=http://gallica.bnf.fr/ark:/12148/bpt6k62424s.image.f192}}.
*{{Citation | last1=Apostol | first1=Tom M. | author1-link=Tom M. Apostol | title=Mathematical analysis | publisher=[[Addison-Wesley]] | edition=2nd | isbn=978-0-201-00288-1 | year=1974}}: §8.14.
* {{citation<!--Deny Citation Bot-->|last=Knopp|first=Konrad|title=Infinite Sequences and Series|location=New York |publisher=Dover Publications|year=1956|isbn=978-0-486-60153-3|bibcode=1956iss..book.....K}}: §3.3, 5.4.
* {{citation|last=Rudin|first=Walter|authorlink=Walter Rudin|title=Principles of Mathematical Analysis|edition=3rd|publisher=McGraw-Hill, Inc.|location=New York|year=1976|isbn=978-0-07-054235-8}}: §3.34.
* {{springer|title=Bertrand criterion|id=p/b015780}}
* {{springer|title=Gauss criterion|id=p/g043420}}
* {{springer|title=Kummer criterion|id=p/k055950}}
* {{citation|last2=Whittaker|first2=E. T.|last1=Watson|first1=G. N.|title=A Course in Modern Analysis|edition=4th|publisher=Cambridge University Press|year=1963|isbn=978-0-521-58807-2}}: §2.36, 2.37.

[[Category:Convergence tests]]
[[Category:Articles containing proofs]]

[[it:Criteri di convergenza#Criterio del rapporto (o di d'Alembert)]]