Ratio test explained

In mathematics, the ratio test is a test (or "criterion") for the convergence of a series

\sum ^{inf ty} _{n =1} a _{n,}

where each term is a real or complex number and is nonzero when 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.

The test

The usual form of the test makes use of the limitThe ratio test states that:

• if L < 1 then the series converges absolutely;
• if L > 1 then the series diverges;
• 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

R = \lim\sup \left| (a _{n+1})/(a _{n}) \right| r = \lim inf \left| (a _{n+1})/(a _{n}) \right| .

Then the ratio test states that:

• if R < 1, the series converges absolutely;
• if r > 1, the series diverges; or equivalently if

\left| (a _{n+1})/(a _{n}) \right|> 1 for all large n (regardless of the value of r), the series also diverges; this is because |a _{n|} is nonzero and increasing and hence does not approach zero;

• the test is otherwise inconclusive.

If the limit L in 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

\sum ^{inf ty (n)/(e ^{n})} _{n =1}

Applying the ratio test, one computes the limit

L = \lim _{n\toinfty} \left| (a _{n+1})/(a _{n}) \right| = \lim _{n\toinfty} \left| ((n +1)/(e ^{n+1}))/((n)/(e ^{n})) \right| = (1)/(e) < 1 .

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

Divergent because L > 1

Consider the series

\sum ^{inf ty (e ^{n})/(n)} _{n =1} .

Putting this into the ratio test:

L = \lim _{n\toinfty} \left| (a _{n+1})/(a _{n}) \right| = \lim _{n\toinfty} \left| ((e ^{n+1})/(n +1))/((e ^{n})/(n)) \right| = e > 1 .

Thus the series diverges.

Inconclusive because L = 1

Consider the three series

\sum ^{inf ty} _{n =1} 1 , \sum ^{inf ty} _{n =1} (1)/(n ^{2}) , \sum ^{inf ty} _{n =1} (( -1 ) ^{n+1})/(n) .

The first series (1 + 1 + 1 + 1 + ⋯) diverges, the second (the one central to the Basel problem) converges absolutely and the third (the alternating harmonic series) converges conditionally. However, the term-by-term magnitude ratios

\left| (a _{n+1})/(a _{n}) \right| of the three series are 1 ,   (n ^{2})/((n +1 ) ^{2})    and   (n)/(n +1) . So, in all three, the limit \lim _{n\toinfty} \left| (a _{n+1})/(a _{n}) \right| is equal to 1. This illustrates that when L = 1, the series may converge or diverge: the ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence.

Proof

Below is a proof of the validity of the original ratio test.

Suppose that

L = \lim _{n\toinfty} \left| (a _{n+1})/(a _{n}) \right| < 1 . 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, consider a real number r such that L < r < 1 . This implies that |a _{n+1} | < r |a _{n} | for sufficiently large n; say, for all n greater than N. Hence |a _{n+i} | < r ^{i|a} _{n} | for each n > N and i > 0, and so\sum ^{inf ty} _{i = N +1} |a _{i|} = \sum ^{inf ty} _{i =1} \left |a _{N+i} \right | < \sum ^{inf ty} _{i =1} r ^{i} |a _{N} | = |a _{N} | \sum ^{inf ty} _{i =1} r ^{i} = |a _{N} | (r)/(1- r) < inf ty.

That is, the series converges absolutely.

On the other hand, if L > 1, then

|a _{n+1} | > |a _{n} | 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.^{[1] [2] [3] [4] [5] [6] [7] [8]}

In all the tests below one assumes that Σa_n is a sum with positive a_n. These tests also may be applied to any series with a finite number of negative terms. Any such series may be written as:

\sum ^{inf ty} _{n =1} a _{n} = \sum ^{N} _{n =1} a _{n+\sum} ^{inf ty} ^{n = N +1} a _{n}

where a_N 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_n) 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_n->∞ρ_n.

All of the tests have regions in which they fail to describe the convergence properties of Σa_n. In fact, no convergence test can fully describe the convergence properties of the series.^{[1] [7]} This is because if Σa_n is convergent, a second convergent series Σb_n can be found which converges more slowly: i.e., it has the property that lim_n->∞ (b_n/a_n) = ∞. Furthermore, if Σa_n is divergent, a second divergent series Σb_n can be found which diverges more slowly: i.e., it has the property that lim_n->∞ (b_n/a_n) = 0. Convergence tests essentially use the comparison test on some particular family of a_n, and fail for sequences which converge or diverge more slowly.

De Morgan hierarchy

Augustus De Morgan proposed a hierarchy of ratio-type tests^{[1] [6]}

The ratio test parameters (

\rho _{n} ) below all generally involve terms of the form D _{n} a _{n/a} _{n+1} - D _{n+1} . This term may be multiplied by a _{n+1} /a _{n} to yield D _{n-D} _{n+1} a _{n+1} /a _{n} . 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:

\rho _{n} \equiv n\left( (a _{n})/(a _{n+1}) -1 \right)

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

The series will:^{[4] [7] [6]}

• Converge when there exists a c>1 such that

\rho _{n} \ge c for all n>N.

• Diverge when

\rho _{n} \le 1 for all n>N.

• Otherwise, the test is inconclusive.

For the limit version, the series will:

• Converge if

\rho = \lim _{n\toinfty} \rho _{n>1} (this includes the case ρ = ∞)

• Diverge if

\lim _{n\toinfty} \rho _{n<1} .

• If ρ = 1, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior.^[1] The series will:

• Converge if

\liminf _{n} \rho _{n} > 1

• Diverge if

\limsup _{n} \rho _{n} < 1

• Otherwise, the test is inconclusive.

Proof of Raabe's test

Defining

\rho _{n} \equiv n\left( (a _{n})/(a _{n+1}) -1 \right) , we need not assume the limit exists; if \limsup\rho _{n<1} , then \sum a _{n} diverges, while if \liminf \rho _{n>1} the sum converges.

The proof proceeds essentially by comparison with

\sum 1 /n ^{R} . Suppose first that \limsup\rho _{n<1} . Of courseif \limsup\rho _{n<0} then a _{n+1} \ge a _{n} for large n , so the sum diverges; assume then that 0 \le\limsup\rho _{n<1} . There exists R< 1 such that \rho _{n\le} R for all n\ge N , which is to say that a _{n} /a _{n+1} \le \left( 1+ (Rn\right)\le)/(e) ^{R/n} . Thus a _{n+1} \ge a ^{- R/n} _{ne} , which implies thata _{n+1} \ge a ^{- R( 1 /N +...+1 /n)} _{Ne} \ge ca ^{- R log (n)} _{Ne} = ca ^{R} _{N/n} for n\ge N ; since R< 1 this shows that \sum a _{n} diverges.

The proof of the other half is entirely analogous, with most of the inequalities simply reversed. We need a preliminary inequality to usein place of the simple

1+ t<e ^{t} that was used above: Fix R and N . Note thatlog \left( 1+ (Rn\right) = (Rn + O\left( (1 {n)/(^{2}\right)}))/())/() . So log \left(\left( 1+ (RN\right) ... \left( 1+ (Rn\right)\right) = R\left( (1 N +...+ (1 n\right) + O( 1 ) = R log (n) + O( 1 ))/())/())/())/() ; hence \left( 1+ (RN\right) ... \left( 1+ (Rn\right)\ge)/(cn))/(^{R}) .

Suppose now that

\liminf\rho _{n>1} . Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there exists R> 1 such that a _{n+1} \le ca ^{- R} _{Nn} for n\ge N ; since R> 1 this shows that \sum a _{n} converges.

3. Bertrand's test

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

Defining:

\rho _{n} \equiv n ln n\left( (a _{n})/(a _{n+1}) -1 \right) -ln n

Bertrand's test^{[1] [7]} asserts that the series will:

• Converge when there exists a c>1 such that

\rho _{n} \ge c for all n>N.

• Diverge when

\rho _{n} \le 1 for all n>N.

• Otherwise, the test is inconclusive.

For the limit version, the series will:

• Converge if

\rho = \lim _{n\toinfty} \rho _{n>1} (this includes the case ρ = ∞)

• Diverge if

\lim _{n\toinfty} \rho _{n<1} .

• If ρ = 1, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior.^{[1] [6]} The series will:

• Converge if

\liminf \rho _{n} > 1

• Diverge if

\limsup \rho _{n} < 1

• Otherwise, the test is inconclusive.

4. Extended Bertrand's test

This extension probably appeared at the first time by Margaret Martin in 1941.^[9] A short proof based on Kummer's test and without technical assumptions (such as existence of the limits, for example) was provided by Vyacheslav Abramov in 2019.^[10]

Let

K\geq 1 be an integer, and let ln _{(K)} (x) denote the K th iterate of natural logarithm, i.e. ln _{(1)} (x) =ln (x) and for any 2 \leq k\leq K , ln _{(k)} (x) =ln _{(k-1)} ( ln (x)) .

Suppose that the ratio

a _{n/a} _{n+1} , when n is large, can be presented in the form(a _{n})/(=1+ a _{n+1}) (1)/(+ n) (1)/(n) \sum ^{K -1} _{i =1} (1)/(+ \prod ^{i ln} _{(k)} (n)) (\rho _{n})/(n\prod ^{K ln} _{(k)} (n)) , K\geq 1 . (The empty sum is assumed to be 0. With K =1 , the test reduces to Bertrand's test.)

The value

\rho _{n} can be presented explicitly in the form\rho _{n} = n\prod ^{K ln} _{(k)} (n)\left( (a _{n})/(a _{n+1}) -1 \right) - \sum ^{j ln} _{(K - k +1 )} (n).

Extended Bertrand's test asserts that the series

• Converge when there exists a

c> 1 such that \rho _{n} \geq c for all n>N .

• Diverge when

\rho _{n} \leq 1 for all n>N .

• Otherwise, the test is inconclusive.

For the limit version, the series

• Converge if

\rho = \lim _{n\toinfty} \rho _{n>1} (this includes the case \rho = inf ty )

• Diverge if

\lim _{n\toinfty} \rho _{n<1} .

• If

\rho = 1 , the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior. The series

• Converge if

\liminf \rho _{n} > 1

• Diverge if

\limsup \rho _{n} < 1

• Otherwise, the test is inconclusive.

For applications of Extended Bertrand's test see birth–death process.

5. Gauss's test

This extension is due to Carl Friedrich Gauss.

Assuming a_n > 0 and r > 1, if a bounded sequence C_n can be found such that for all n:^{[2] [4] [6] [7]}

(a _{n})/(=1+ a _{n+1}) (\rho)/(+ n) (C _{n})/(n ^{r})

then the series will:

• Converge if

\rho> 1

• Diverge if

\rho \le 1

6. Kummer's test

This extension is due to Ernst Kummer.

Let ζ_n be an auxiliary sequence of positive constants. Define

\rho _{n} \equiv \left(\zeta _{n} (a _{n})/(a _{n+1}) - \zeta _{n+1} \right)

Kummer's test states that the series will:^{[2] [3] [7] [8]}

• Converge if there exists a

c> 0 such that \rho _{n} \ge c for all n>N. (Note this is not the same as saying \rho _{n} > 0 )

• Diverge if

\rho _{n} \le 0 for all n>N and \sum ^{inf ty} _{n =1} 1 /\zeta _{n} diverges.

For the limit version, the series will:^{[4] [6]}

• Converge if

\lim _{n\toinfty} \rho _{n>0} (this includes the case ρ = ∞)

• Diverge if

\lim _{n\toinfty} \rho _{n<0} and \sum ^{inf ty} _{n =1} 1 /\zeta _{n} diverges.

• Otherwise the test is inconclusive

When the above limit does not exist, it may be possible to use limits superior and inferior.^[1] The series will

• Converge if

\liminf _{n} \rho _{n} > 0

• Diverge if

\limsup _{n} \rho _{n} < 0 and \sum 1 /\zeta _{n} 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:^[1]

• For the ratio test, let ζ_n=1. Then:

\rho _{Kummer} = \left( (a _{n})/(a _{n+1}) -1 \right) = 1 /\rho _{Ratio-1}

• For Raabe's test, let ζ_n=n. Then:

\rho _{Kummer} = \left(n (a _{n})/(a _{n+1}) - (n +1 )\right) = \rho _{Raabe-1}

• For Bertrand's test, let ζ_n=n ln(n). Then:

\rho _{Kummer} = n ln (n)\left( (a _{n})/(a _{n+1}) \right) - (n +1 ) ln (n +1 )

Using

ln (n +1 ) =ln (n) +ln ( 1+1 /n) and approximating ln ( 1+1 /n) → 1 /n for large n, which is negligible compared to the other terms, \rho _{Kummer} may be written:\rho _{Kummer} = n ln (n)\left( (a _{n})/(a _{n+1}) -1 \right) -ln (n) -1 = \rho _{Bertrand-1}

• For Extended Bertrand's test, let

ln _{(k)} (n +1 ) =ln _{(k)} (n) + (1)/(+ O\left( n\prod ^{k -1} _{j =1} ln _{(j)} (n)) (1)/(n ^{2}) \right), where the empty product is assumed to be 1. Then,\rho _{Kummer} = n\prod ^{K ln} _{(k)} (n) (a _{n})/(a _{n+1}) - (n +1 )\left[\prod ^{K\left( ln} _{(k)} (n) + (1)/(n\prod ^{k -1} _{j =1} ln _{(j)} (n)) \right)\right] + o( 1 ) = n\prod ^{K ln} _{(k)} (n)\left( (a _{n})/(a _{n+1}) -1 \right) - \sum ^{j ln} _{(K - k +1 )} (n) -1+ o( 1 ).

Hence,

\rho _{Kummer} = \rho _{ExtendedBertrand-1.}

Note that for these four tests, the higher they are in the De Morgan hierarchy, the more slowly the

1 /\zeta _{n} series diverges.

Proof of Kummer's test

If

\rho _{n>0} then fix a positive number 0 <\delta<\rho _{n} . There existsa natural number N such that for every n>N, \delta\leq\zeta _{n} (a _{n})/(a _{n+1}) - \zeta _{n+1} . Since a _{n+1} > 0 , for every n> N, 0 \leq \delta a _{n+1} \leq \zeta _{n} a _{n} - \zeta _{n+1} a _{n+1} . In particular \zeta _{n+1} a _{n+1} \leq \zeta _{n} a _{n} for all n\geq N which means that starting from the indexN the sequence \zeta _{n} a _{n} > 0 is monotonically decreasing andpositive which in particular implies that it is bounded below by 0. Therefore, the limit\lim _{n\toinfty} \zeta _{n} a _{n} = L exists.This implies that the positive telescoping series\sum ^{inf ty} _{n =1} \left(\zeta _{n} a _{n} - \zeta _{n+1} a _{n+1} \right) is convergent,and since for all n>N, \delta a _{n+1} \leq \zeta _{n} a _{n} - \zeta _{n+1} a _{n+1} by the direct comparison test for positive series, the series\sum ^{inf ty} _{n =1} \delta a _{n+1} is convergent.

On the other hand, if

\rho< 0 , then there is an N such that \zeta _{n} a _{n} is increasing for n>N . In particular, there exists an \epsilon> 0 for which \zeta _{n} a _{n>\epsilon} for all n>N , and so \sum _{n} a _{n=\sum} _{n} (a _{n\zeta} _{n})/(\zeta _{n}) diverges by comparison with \sum _{n} (\epsilon)/(\zeta _{n}) .

Tong's modification of Kummer's test

A new version of Kummer's test was established by Tong.^[3] See also ^{[5] [8] [11]} for further discussions and new proofs. The provided modification of Kummer's theorem characterizesall positive series, and the convergence or divergence can be formulated in the form of two necessary and sufficient conditions, one for convergence and another for divergence.

• Series

\sum ^{inf ty} _{n =1} a _{n} converges if and only if there exists a positive sequence \zeta _{n} , n =1 , 2 , ... , such that \zeta _{n (a _{n})/(a _{n+1})} - \zeta _{n+1} \geq c> 0 .

• Series

\sum ^{inf ty} _{n =1} a _{n} diverges if and only if there exists a positive sequence \zeta _{n} , n =1 , 2 , ... , such that \zeta _{n (a _{n})/(a _{n+1})} - \zeta _{n+1} \leq 0 , and \sum ^{inf ty} _{n =1} (1)/(\zeta _{n}) =inf ty.

The first of these statements can be simplified as follows: ^[12]

• Series

\sum ^{inf ty} _{n =1} a _{n} converges if and only if there exists a positive sequence \zeta _{n} , n =1 , 2 , ... , such that \zeta _{n (a _{n})/(a _{n+1})} - \zeta _{n+1} =1 .

The second statement can be simplified similarly:

• Series

\sum ^{inf ty} _{n =1} a _{n} diverges if and only if there exists a positive sequence \zeta _{n} , n =1 , 2 , ... , such that \zeta _{n (a _{n})/(a _{n+1})} - \zeta _{n+1} =0 , and \sum ^{inf ty} _{n =1} (1)/(\zeta _{n}) =inf ty.

However, it becomes useless, since the condition

\sum ^{inf ty} _{n =1} (1)/(\zeta _{n}) =inf ty in this case reduces to the original claim \sum ^{inf ty} _{n =1} a _{n=infty.}

Frink's ratio test

Another ratio test that can be set in the framework of Kummer's theorem was presented by Orrin Frink^[13] 1948.

Suppose

a _{n} is a sequence in C \setminus\{ 0 \} ,

• If

\limsup _{n → infty} ( (|a _{n+1} |)/(|a _{n|}) ) ^{n< (1 e)/()} , then the series \sum _{na} _{n} converges absolutely.

• If there is

N\in N such that ( (|a _{n+1} |)/(|a _{n|}) ) ^{n\geq (1 e)/()} for all n\geq N , then \sum _{n|a} _{n|} diverges.

This result reduces to a comparison of

\sum _{n|a} _{n|} with a power series \sum _{n} n ^{-p} , and can be seen to be related to Raabe's test.^[14]

Ali's second ratio test

A more refined ratio test is the second ratio test:^{[4] [6]} For

a _{n>0} define:

L _{0\equiv\lim} _{n →} (a _{2n})/(a _{n})
L _{1\equiv\lim} _{n →} (a _{2n+1})/(a _{n})
L\equiv max (L _{0,L} _{1)}

By the second ratio test, the series will:

• Converge if

L< (1)/(2)

• Diverge if

L> (1)/(2)

• If

L = (1)/(2) then the test is inconclusive.

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

L _{0\equiv\limsup} _{n →} (a _{2n})/(a _{n})		L _{1\equiv\limsup} _{n →} (a _{2n+1})/(a _{n})
\ell _{0\equiv\liminf} _{n →} (a _{2n})/(a _{n})		\ell _{1\equiv\liminf} _{n →} (a _{2n+1})/(a _{n})
L\equiv max (L _{0,L} _{1)}		\ell\equiv min (\ell _{0,\ell} _{1)}

Then the series will:

• Converge if

L< (1)/(2)

• Diverge if

\ell> (1)/(2)

• If

\ell \le (1)/(2) \le L then the test is inconclusive.

Ali's mth ratio test

This test is a direct extension of the second ratio test.^{[4] [6]} For

0 \leq k\leq m -1 , and positive a _{n} define:

L _{k\equiv\lim} _{n →} (a _{mn+k})/(a _{n})
L\equiv max (L _{0,L} _{1,\ldots,L} _{m-1} )

By the m th ratio test, the series will:

• Converge if

L< (1)/(m)

• Diverge if

L> (1)/(m)

• If

L = (1)/(m) then the test is inconclusive.

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

0 \leq k\leq m -1 define:

L _{k\equiv\limsup} _{n →} (a _{mn+k})/(a _{n})
\ell _{k\equiv\liminf} _{n →} (a _{mn+k})/(a _{n})
L\equiv max (L _{0,L} _{1,\ldots,L} _{m-1} )		\ell\equiv min (\ell _{0,\ell} _{1,\ldots,\ell} _{m-1} )

Then the series will:

• Converge if

L< (1)/(m)

• Diverge if

\ell> (1)/(m)

• If

\ell \leq (1)/(m) \leq L , then the test is inconclusive.

Ali--Deutsche Cohen φ-ratio test

This test is an extension of the

m th ratio test.^[15]

Assume that the sequence

a _{n} is a positive decreasing sequence.

Let

\varphi: Z ^{+\toZ} ^{+} be such that \lim _{n\toinfty} (n)/(\varphi(n)) exists. Denote \alpha = \lim _{n\toinfty} (n)/(\varphi(n)) , and assume 0 <\alpha< 1 .

Assume also that

\lim _{n\toinfty} (a _{\varphi(n)})/(a _{n}) = L.

Then the series will:

• Converge if

L<\alpha

• Diverge if

L>\alpha

• If

L = \alpha , then the test is inconclusive.

See also

• Root test
• Radius of convergence

References

• .

§8.14.

• : §3.3, 5.4.
• : §3.34.
  • • • : §2.36, 2.37.

Notes and References

1. Book: Bromwich, T. J. I'A . 1908 . An Introduction To The Theory of Infinite Series . Merchant Books . Thomas John I'Anson Bromwich.
2. Book: Knopp, Konrad . 1954 . Theory and Application of Infinite Series . London . Blackie & Son Ltd. . Konrad Knopp .
3. Tong . Jingcheng. May 1994. Kummer's Test Gives Characterizations for Convergence or Divergence of all Positive Series. 2974907. The American Mathematical Monthly . 101 . 5 . 450–452 . 10.2307/2974907 .
4. Ali . Sayel A.. 2008 . The mth Ratio Test: New Convergence Test for Series . The American Mathematical Monthly . 115 . 6 . 514–524 . 10.1080/00029890.2008.11920558. 16336333. 21 November 2018 .
5. Samelson . Hans. November 1995. More on Kummer's Test. 2974510. The American Mathematical Monthly . 102 . 9 . 817–818 . 10.2307/2974510 .
6. Web site: The mth Ratio Convergence Test and Other Unconventional Convergence Tests. Blackburn. Kyle . 4 May 2012 . University of Washington College of Arts and Sciences . 27 November 2018 .
7. Ďuriš . František . 2009 . Infinite series: Convergence tests . Bachelor's thesis . Katedra Informatiky, Fakulta Matematiky, Fyziky a Informatiky, Univerzita Komenského, Bratislava . 28 November 2018 .
8. Ďuriš . František . 2 February 2018 . On Kummer's test of convergence and its relation to basic comparison tests . 1612.05167 . math.HO .
9. A sequence of limit tests for the convergence of series . Martin . Margaret . 1941 . Bulletin of the American Mathematical Society . 47 . 6. 452–457 . 10.1090/S0002-9904-1941-07477-X . free .
10. Abramov . Vyacheslav M. . May 2020 . Extension of the Bertrand–De Morgan test and its application . The American Mathematical Monthly . 127 . 5 . 444–448 . 10.1080/00029890.2020.1722551 . 1901.05843 . 199552015 .
11. Abramov . Vyacheslav, M. . 21 June 2021 . A simple proof of Tong's theorem . 2106.13808 . math.HO .
12. Abramov . Vyacheslav M.. May 2022. Evaluating the sum of convergent positive series. Publications de l'Institut Mathématique . Nouvelle Série . 111 . 125 . 41–53 . 10.2298/PIM2225041A. 237499616 .
13. Frink . Orrin . October 1948. A ratio test . Bulletin of the American Mathematical Society . 54 . 10 . 953-953.
14. 1949. Stark . Marceli . On the ratio test of Frink . Colloquium Mathematicum . 2 . 1 . 46-47 .
15. phi-ratio tests . Ali . Sayel . Cohen . Marion Deutsche . 2012 . Elemente der Mathematik . 67 . 4. 164–168 . 10.4171/EM/206 . free .