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Ratio test

From Wikipedia, the free encyclopedia
Criterion for the convergence of a series
Part of a series of articles about
Calculus
∫abf′(t)dt=f(b)−f(a){\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}

Inmathematics, theratio test is atest (or "criterion") for theconvergence of aseries

∑n=1∞an,{\displaystyle \sum _{n=1}^{\infty }a_{n},}

where each term is areal orcomplex number andan is nonzero whenn is large. The test was first published byJean le Rond d'Alembert and is sometimes known asd'Alembert's ratio testor as theCauchy ratio test.[1]

The test

[edit]
Decision diagram for the ratio test

The usual form of the test makes use of thelimit

L=limn→∞|an+1an|.{\displaystyle L=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|.}1

The ratio test states that:

  • ifL < 1 then the seriesconverges absolutely;
  • ifL > 1 then the seriesdiverges;
  • ifL = 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 limitL fails to exist, iflimit superior andlimit inferior are used. The test criteria can also be refined so that the test is sometimes conclusive even whenL = 1. More specifically, let

R=limsup|an+1an|{\displaystyle R=\lim \sup \left|{\frac {a_{n+1}}{a_{n}}}\right|}
r=liminf|an+1an|{\displaystyle r=\lim \inf \left|{\frac {a_{n+1}}{a_{n}}}\right|}.

Then the ratio test states that:[2][3]

If the limitL in (1) exists, we must haveL =R =r. So the original ratio test is a weaker version of the refined one.

Examples

[edit]

Convergent becauseL < 1

[edit]

Consider the series

∑n=1∞nen{\displaystyle \sum _{n=1}^{\infty }{\frac {n}{e^{n}}}}

Applying the ratio test, one computes the limit

L=limn→∞|an+1an|=limn→∞|n+1en+1nen|=1e<1.{\displaystyle 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.}

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

Divergent becauseL > 1

[edit]

Consider the series

∑n=1∞enn.{\displaystyle \sum _{n=1}^{\infty }{\frac {e^{n}}{n}}.}

Putting this into the ratio test:

L=limn→∞|an+1an|=limn→∞|en+1n+1enn|=e>1.{\displaystyle 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.}

Thus the series diverges.

Inconclusive becauseL = 1

[edit]

Consider the three series

∑n=1∞1,{\displaystyle \sum _{n=1}^{\infty }1,}
∑n=1∞1n2,{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}},}
∑n=1∞(−1)n+1n.{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}.}

The first series (1 + 1 + 1 + 1 + ⋯) diverges, the second (the one central to theBasel problem) converges absolutely and the third (thealternating harmonic series) converges conditionally. However, the term-by-term magnitude ratios|an+1an|{\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|} of the three series are1,{\displaystyle 1,}   n2(n+1)2{\displaystyle {\frac {n^{2}}{(n+1)^{2}}}}    and   nn+1{\displaystyle {\frac {n}{n+1}}}. So, in all three, the limitlimn→∞|an+1an|{\displaystyle \lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|} is equal to 1. This illustrates that whenL = 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

[edit]
In this example, the ratio of adjacent terms in the blue sequence converges to L=1/2. We chooser = (L+1)/2 = 3/4. Then the blue sequence is dominated by the red sequencerk for alln ≥ 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 thatr=lim infn→∞|an+1an|>1{\displaystyle r=\liminf _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|>1}. We also suppose that(an){\displaystyle (a_{n})} has infinite non-zero members, otherwise the series is just a finite sum hence it converges. Then there exists someℓ∈(1;r){\displaystyle \ell \in (1;r)} such that there exists a natural numbern0≥2{\displaystyle n_{0}\geq 2} satisfyingan0≠0{\displaystyle a_{n_{0}}\neq 0} and|an+1an|>ℓ{\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|>\ell } for alln≥n0{\displaystyle n\geq n_{0}}, because if no suchℓ{\displaystyle \ell } exists then there exists arbitrarily largen{\displaystyle n} satisfying|an+1an|<ℓ{\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|<\ell } for everyℓ∈(1;r){\displaystyle \ell \in (1;r)}, then we can find a subsequence(ank)k=1∞{\displaystyle \left(a_{n_{k}}\right)_{k=1}^{\infty }} satisfyinglim supn→∞|ank+1ank|≤ℓ<r{\displaystyle \limsup _{n\to \infty }\left|{\frac {a_{n_{k}+1}}{a_{n_{k}}}}\right|\leq \ell <r}, but this contradicts the fact thatr{\displaystyle r} is thelimit inferior of|an+1an|{\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|} asn→∞{\displaystyle n\to \infty }, implying the existence ofℓ{\displaystyle \ell }. Then we notice that forn≥n0+1{\displaystyle n\geq n_{0}+1},|an|>ℓ|an−1|>ℓ2|an−2|>...>ℓn−n0|an0|{\displaystyle |a_{n}|>\ell |a_{n-1}|>\ell ^{2}|a_{n-2}|>...>\ell ^{n-n_{0}}\left|a_{n_{0}}\right|}. Notice thatℓ>1{\displaystyle \ell >1} soℓn→∞{\displaystyle \ell ^{n}\to \infty } asn→∞{\displaystyle n\to \infty } and|an0|>0{\displaystyle \left|a_{n_{0}}\right|>0}, this implies(an){\displaystyle (a_{n})} diverges so the series∑n=1∞an{\displaystyle \sum _{n=1}^{\infty }a_{n}} diverges by then-th term test.
Now supposeR=lim supn→∞|an+1an|<1{\displaystyle R=\limsup _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|<1}. Similar to the above case, we may find a natural numbern1{\displaystyle n_{1}} and ac∈(R;1){\displaystyle c\in (R;1)} such that|an|≤cn−n1|an1|{\displaystyle |a_{n}|\leq c^{n-n_{1}}\left|a_{n_{1}}\right|} forn≥n1{\displaystyle n\geq n_{1}}. Then∑n=1∞|an|=∑k=1n1−1|ak|+∑n=n1∞|an|≤∑k=1n1−1|ak|+∑n=n1∞cn−n1|an1|=∑k=1n1−1|ak|+|an1|∑n=0∞cn.{\displaystyle \sum _{n=1}^{\infty }|a_{n}|=\sum _{k=1}^{n_{1}-1}|a_{k}|+\sum _{n=n_{1}}^{\infty }|a_{n}|\leq \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}.}The series∑n=0∞cn{\displaystyle \sum _{n=0}^{\infty }c^{n}} is thegeometric series with common ratioc∈(0;1){\displaystyle c\in (0;1)}, hence∑n=0∞cn=11−c{\displaystyle \sum _{n=0}^{\infty }c^{n}={\frac {1}{1-c}}} which is finite. The sum∑k=1n1−1|ak|{\displaystyle \sum _{k=1}^{n_{1}-1}|a_{k}|} is a finite sum and hence it is bounded, this implies the series∑n=1∞|an|{\displaystyle \sum _{n=1}^{\infty }|a_{n}|} converges by themonotone convergence theorem and the series∑n=1∞an{\displaystyle \sum _{n=1}^{\infty }a_{n}} converges by the absolute convergence test.
When the limit|an+1an|{\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|} exists and equals toL{\displaystyle L} thenr=R=L{\displaystyle r=R=L}, this gives the original ratio test.

Extensions forL = 1

[edit]

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 allow one to deal with this case.[4][5][6][7][8][9][10][11]

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

∑n=1∞an=∑n=1Nan+∑n=N+1∞an{\displaystyle \sum _{n=1}^{\infty }a_{n}=\sum _{n=1}^{N}a_{n}+\sum _{n=N+1}^{\infty }a_{n}}

whereaN 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 atn=1.

Each test defines a test parameter (ρn) 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 limn->∞ρn.

All of the tests have regions in which they fail to describe the convergence properties of Σan. In fact, no convergence test can fully describe the convergence properties of the series.[4][10] This is because if Σan is convergent, a second convergent series Σbn can be found which converges more slowly: i.e., it has the property that limn->∞ (bn/an) = ∞. Furthermore, if Σan is divergent, a second divergent series Σbn can be found which diverges more slowly: i.e., it has the property that limn->∞ (bn/an) = 0. Convergence tests essentially use the comparison test on some particular family of an, and fail for sequences which converge or diverge more slowly.

De Morgan hierarchy

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Augustus De Morgan proposed a hierarchy of ratio-type tests[4][9]

The ratio test parameters (ρn{\displaystyle \rho _{n}}) below all generally involve terms of the formDnan/an+1−Dn+1{\displaystyle D_{n}a_{n}/a_{n+1}-D_{n+1}}. This term may be multiplied byan+1/an{\displaystyle a_{n+1}/a_{n}} to yieldDn−Dn+1an+1/an{\displaystyle 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

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

2. Raabe's test

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This extension is due toJoseph Ludwig Raabe. Define:

ρn≡n(anan+1−1){\displaystyle \rho _{n}\equiv n\left({\frac {a_{n}}{a_{n+1}}}-1\right)}

(and some extra terms, see Ali, Blackburn, Feld, Duris (none), Duris2)[clarification needed]

The series will:[7][10][9]

For the limit version,[12] the series will:

When the above limit does not exist, it may be possible to use limits superior and inferior.[4] The series will:

Proof of Raabe's test
[edit]

Definingρn≡n(anan+1−1){\displaystyle \rho _{n}\equiv n\left({\frac {a_{n}}{a_{n+1}}}-1\right)}, we need not assume the limit exists; iflim supρn<1{\displaystyle \limsup \rho _{n}<1}, then∑an{\displaystyle \sum a_{n}} diverges, while iflim infρn>1{\displaystyle \liminf \rho _{n}>1} the sum converges.

The proof proceeds essentially by comparison with∑1/nR{\displaystyle \sum 1/n^{R}}. Suppose first thatlim supρn<1{\displaystyle \limsup \rho _{n}<1}. Of courseiflim supρn<0{\displaystyle \limsup \rho _{n}<0} thenan+1≥an{\displaystyle a_{n+1}\geq a_{n}} for largen{\displaystyle n}, so the sum diverges; assume then that0≤lim supρn<1{\displaystyle 0\leq \limsup \rho _{n}<1}. There existsR<1{\displaystyle R<1} such thatρn≤R{\displaystyle \rho _{n}\leq R} for alln≥N{\displaystyle n\geq N}, which is to say thatan/an+1≤(1+Rn)≤eR/n{\displaystyle a_{n}/a_{n+1}\leq \left(1+{\frac {R}{n}}\right)\leq e^{R/n}}. Thusan+1≥ane−R/n{\displaystyle a_{n+1}\geq a_{n}e^{-R/n}}, which implies thatan+1≥aNe−R(1/N+⋯+1/n)≥caNe−Rlog⁡(n)=caN/nR{\displaystyle a_{n+1}\geq a_{N}e^{-R(1/N+\dots +1/n)}\geq ca_{N}e^{-R\log(n)}=ca_{N}/n^{R}} forn≥N{\displaystyle n\geq N}; sinceR<1{\displaystyle R<1} this shows that∑an{\displaystyle \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 simple1+t<et{\displaystyle 1+t<e^{t}} that was used above: FixR{\displaystyle R} andN{\displaystyle N}. Note thatlog⁡(1+Rn)=Rn+O(1n2){\displaystyle \log \left(1+{\frac {R}{n}}\right)={\frac {R}{n}}+O\left({\frac {1}{n^{2}}}\right)}. Solog⁡((1+RN)…(1+Rn))=R(1N+⋯+1n)+O(1)=Rlog⁡(n)+O(1){\displaystyle \log \left(\left(1+{\frac {R}{N}}\right)\dots \left(1+{\frac {R}{n}}\right)\right)=R\left({\frac {1}{N}}+\dots +{\frac {1}{n}}\right)+O(1)=R\log(n)+O(1)}; hence(1+RN)…(1+Rn)≥cnR{\displaystyle \left(1+{\frac {R}{N}}\right)\dots \left(1+{\frac {R}{n}}\right)\geq cn^{R}}.

Suppose now thatlim infρn>1{\displaystyle \liminf \rho _{n}>1}. Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there existsR>1{\displaystyle R>1} such thatan+1≤caNn−R{\displaystyle a_{n+1}\leq ca_{N}n^{-R}} forn≥N{\displaystyle n\geq N}; sinceR>1{\displaystyle R>1} this shows that∑an{\displaystyle \sum a_{n}} converges.

(A faster approach to proving divergence:an/an+1≤(1+Rn)≤(1+1n){\displaystyle a_{n}/a_{n+1}\leq \left(1+{\frac {R}{n}}\right)\leq \left(1+{\frac {1}{n}}\right)}. Thusann≤an+1(n+1){\displaystyle a_{n}n\leq a_{n+1}(n+1)}, which implies thatann{\displaystyle a_{n}n} is monotone increasing forn≥N{\displaystyle n\geq N}; sincean>0{\displaystyle a_{n}>0}, there must exists a constantϵ>0{\displaystyle \epsilon >0} such thatann>ϵ{\displaystyle a_{n}n>\epsilon } for alln≥N{\displaystyle n\geq N}. Therefore,an≥ϵn{\displaystyle a_{n}\geq {\frac {\epsilon }{n}}} and∑an{\displaystyle \sum a_{n}} is diverges.)

3. Bertrand's test

[edit]

This extension is due toJoseph Bertrand andAugustus De Morgan.

Defining:

ρn≡nln⁡n(anan+1−1)−ln⁡n{\displaystyle \rho _{n}\equiv n\ln n\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\ln n}

Bertrand's test[4][10] asserts that the series will:

For the limit version, the series will:

When the above limit does not exist, it may be possible to use limits superior and inferior.[4][9][13] The series will:

4. Extended Bertrand's test

[edit]

This extension probably appeared at the first time by Margaret Martin in 1941.[14] 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.[15]

LetK≥1{\displaystyle K\geq 1} be an integer, and letln(K)⁡(x){\displaystyle \ln _{(K)}(x)} denote theK{\displaystyle K}thiterate ofnatural logarithm, i.e.ln(1)⁡(x)=ln⁡(x){\displaystyle \ln _{(1)}(x)=\ln(x)} and for any2≤k≤K{\displaystyle 2\leq k\leq K},ln(k)⁡(x)=ln(k−1)⁡(ln⁡(x)){\displaystyle \ln _{(k)}(x)=\ln _{(k-1)}(\ln(x))}.

Suppose that the ratioan/an+1{\displaystyle a_{n}/a_{n+1}}, whenn{\displaystyle n} is large, can be presented in the form

anan+1=1+1n+1n∑i=1K−11∏k=1iln(k)⁡(n)+ρnn∏k=1Kln(k)⁡(n),K≥1.{\displaystyle {\frac {a_{n}}{a_{n+1}}}=1+{\frac {1}{n}}+{\frac {1}{n}}\sum _{i=1}^{K-1}{\frac {1}{\prod _{k=1}^{i}\ln _{(k)}(n)}}+{\frac {\rho _{n}}{n\prod _{k=1}^{K}\ln _{(k)}(n)}},\quad K\geq 1.}

(The empty sum is assumed to be 0. WithK=1{\displaystyle K=1}, the test reduces to Bertrand's test.)

The valueρn{\displaystyle \rho _{n}} can be presented explicitly in the form

ρn=n∏k=1Kln(k)⁡(n)(anan+1−1)−∑j=1K∏k=1jln(K−k+1)⁡(n).{\displaystyle \rho _{n}=n\prod _{k=1}^{K}\ln _{(k)}(n)\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\sum _{j=1}^{K}\prod _{k=1}^{j}\ln _{(K-k+1)}(n).}

Extended Bertrand's test asserts that the series

For the limit version, the series

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

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

5. Gauss's test

[edit]

This extension is due toCarl Friedrich Gauss.

Assumingan > 0 andr > 1, if a bounded sequenceCn can be found such that for alln:[5][7][9][10]

anan+1=1+ρn+Cnnr{\displaystyle {\frac {a_{n}}{a_{n+1}}}=1+{\frac {\rho }{n}}+{\frac {C_{n}}{n^{r}}}}

then the series will:

6. Kummer's test

[edit]

This extension is due toErnst Kummer.

Let ζn be an auxiliary sequence of positive constants. Define

ρn≡(ζnanan+1−ζn+1){\displaystyle \rho _{n}\equiv \left(\zeta _{n}{\frac {a_{n}}{a_{n+1}}}-\zeta _{n+1}\right)}

Kummer's test states that the series will:[5][6][10][11]

For the limit version, the series will:[16][7][9]

When the above limit does not exist, it may be possible to use limits superior and inferior.[4] The series will

Special cases
[edit]

All of the tests in De Morgan's hierarchy except Gauss's test can easily be seen as special cases of Kummer's test:[4]

  • For the ratio test, let ζn=1. Then:
ρKummer=(anan+1−1)=1/ρRatio−1{\displaystyle \rho _{\text{Kummer}}=\left({\frac {a_{n}}{a_{n+1}}}-1\right)=1/\rho _{\text{Ratio}}-1}
  • For Raabe's test, let ζn=n. Then:
ρKummer=(nanan+1−(n+1))=ρRaabe−1{\displaystyle \rho _{\text{Kummer}}=\left(n{\frac {a_{n}}{a_{n+1}}}-(n+1)\right)=\rho _{\text{Raabe}}-1}
  • For Bertrand's test, let ζn=n ln(n). Then:
ρKummer=nln⁡(n)(anan+1)−(n+1)ln⁡(n+1){\displaystyle \rho _{\text{Kummer}}=n\ln(n)\left({\frac {a_{n}}{a_{n+1}}}\right)-(n+1)\ln(n+1)}
Usingln⁡(n+1)=ln⁡(n)+ln⁡(1+1/n){\displaystyle \ln(n+1)=\ln(n)+\ln(1+1/n)} andapproximatingln⁡(1+1/n)→1/n{\displaystyle \ln(1+1/n)\rightarrow 1/n} for largen, which is negligible compared to the other terms,ρKummer{\displaystyle \rho _{\text{Kummer}}} may be written:
ρKummer=nln⁡(n)(anan+1−1)−ln⁡(n)−1=ρBertrand−1{\displaystyle \rho _{\text{Kummer}}=n\ln(n)\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\ln(n)-1=\rho _{\text{Bertrand}}-1}
ln(k)⁡(n+1)=ln(k)⁡(n)+1n∏j=1k−1ln(j)⁡(n)+O(1n2),{\displaystyle \ln _{(k)}(n+1)=\ln _{(k)}(n)+{\frac {1}{n\prod _{j=1}^{k-1}\ln _{(j)}(n)}}+O\left({\frac {1}{n^{2}}}\right),}

where the empty product is assumed to be 1. Then,

ρKummer=n∏k=1Kln(k)⁡(n)anan+1−(n+1)[∏k=1K(ln(k)⁡(n)+1n∏j=1k−1ln(j)⁡(n))]+o(1)=n∏k=1Kln(k)⁡(n)(anan+1−1)−∑j=1K∏k=1jln(K−k+1)⁡(n)−1+o(1).{\displaystyle \rho _{\text{Kummer}}=n\prod _{k=1}^{K}\ln _{(k)}(n){\frac {a_{n}}{a_{n+1}}}-(n+1)\left[\prod _{k=1}^{K}\left(\ln _{(k)}(n)+{\frac {1}{n\prod _{j=1}^{k-1}\ln _{(j)}(n)}}\right)\right]+o(1)=n\prod _{k=1}^{K}\ln _{(k)}(n)\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\sum _{j=1}^{K}\prod _{k=1}^{j}\ln _{(K-k+1)}(n)-1+o(1).}

Hence,

ρKummer=ρExtended Bertrand−1.{\displaystyle \rho _{\text{Kummer}}=\rho _{\text{Extended Bertrand}}-1.}

Note that for these four tests, the higher they are in the De Morgan hierarchy, the more slowly the1/ζn{\displaystyle 1/\zeta _{n}} series diverges.

Proof of Kummer's test
[edit]

Ifρn>0{\displaystyle \rho _{n}>0} then fix a positive number0<δ<ρn{\displaystyle 0<\delta <\rho _{n}}. There existsa natural numberN{\displaystyle N} such that for everyn>N,{\displaystyle n>N,}

δ≤ζnanan+1−ζn+1.{\displaystyle \delta \leq \zeta _{n}{\frac {a_{n}}{a_{n+1}}}-\zeta _{n+1}.}

Sincean+1>0{\displaystyle a_{n+1}>0}, for everyn>N,{\displaystyle n>N,}

0≤δan+1≤ζnan−ζn+1an+1.{\displaystyle 0\leq \delta a_{n+1}\leq \zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}.}

In particularζn+1an+1≤ζnan{\displaystyle \zeta _{n+1}a_{n+1}\leq \zeta _{n}a_{n}} for alln≥N{\displaystyle n\geq N} which means that starting from the indexN{\displaystyle N}the sequenceζnan>0{\displaystyle \zeta _{n}a_{n}>0} is monotonically decreasing andpositive which in particular implies that it is bounded below by 0. Therefore, the limit

limn→∞ζnan=L{\displaystyle \lim _{n\to \infty }\zeta _{n}a_{n}=L} exists.

This implies that the positivetelescoping series

∑n=1∞(ζnan−ζn+1an+1){\displaystyle \sum _{n=1}^{\infty }\left(\zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}\right)} is convergent,

and since for alln>N,{\displaystyle n>N,}

δan+1≤ζnan−ζn+1an+1{\displaystyle \delta a_{n+1}\leq \zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}}

by thedirect comparison test for positive series, the series∑n=1∞δan+1{\displaystyle \sum _{n=1}^{\infty }\delta a_{n+1}} is convergent.

On the other hand, ifρ<0{\displaystyle \rho <0}, then there is anN such thatζnan{\displaystyle \zeta _{n}a_{n}} is increasing forn>N{\displaystyle n>N}. In particular, there exists anϵ>0{\displaystyle \epsilon >0} for whichζnan>ϵ{\displaystyle \zeta _{n}a_{n}>\epsilon } for alln>N{\displaystyle n>N}, and so∑nan=∑nanζnζn{\displaystyle \sum _{n}a_{n}=\sum _{n}{\frac {a_{n}\zeta _{n}}{\zeta _{n}}}} diverges by comparison with∑nϵζn{\displaystyle \sum _{n}{\frac {\epsilon }{\zeta _{n}}}}.

Tong's modification of Kummer's test

[edit]

A new version of Kummer's test was established by Tong.[6] See also[8][11][17]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.

The first of these statements can be simplified as follows:[18]

The second statement can be simplified similarly:

However, it becomes useless, since the condition∑n=1∞1ζn=∞{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\zeta _{n}}}=\infty } in this case reduces to the original claim∑n=1∞an=∞.{\displaystyle \sum _{n=1}^{\infty }a_{n}=\infty .}

Frink's ratio test

[edit]

Another ratio test that can be set in the framework of Kummer's theorem was presented byOrrin Frink[19] 1948.

Supposean{\displaystyle a_{n}} is a sequence inC∖{0}{\displaystyle \mathbb {C} \setminus \{0\}},

This result reduces to a comparison of∑n|an|{\displaystyle \sum _{n}|a_{n}|} with apower series∑nn−p{\displaystyle \sum _{n}n^{-p}}, and can be seen to be related to Raabe's test.[20]

Ali's second ratio test

[edit]

A more refined ratio test is the second ratio test:[7][9]Foran>0{\displaystyle a_{n}>0} define:

L0≡limn→∞a2nan{\displaystyle L_{0}\equiv \lim _{n\rightarrow \infty }{\frac {a_{2n}}{a_{n}}}}
L1≡limn→∞a2n+1an{\displaystyle L_{1}\equiv \lim _{n\rightarrow \infty }{\frac {a_{2n+1}}{a_{n}}}}
L≡max(L0,L1){\displaystyle L\equiv \max(L_{0},L_{1})}

By the second ratio test, the series will:

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

L0≡lim supn→∞a2nan{\displaystyle L_{0}\equiv \limsup _{n\rightarrow \infty }{\frac {a_{2n}}{a_{n}}}}L1≡lim supn→∞a2n+1an{\displaystyle L_{1}\equiv \limsup _{n\rightarrow \infty }{\frac {a_{2n+1}}{a_{n}}}}
ℓ0≡lim infn→∞a2nan{\displaystyle \ell _{0}\equiv \liminf _{n\rightarrow \infty }{\frac {a_{2n}}{a_{n}}}}ℓ1≡lim infn→∞a2n+1an{\displaystyle \ell _{1}\equiv \liminf _{n\rightarrow \infty }{\frac {a_{2n+1}}{a_{n}}}}
L≡max(L0,L1){\displaystyle L\equiv \max(L_{0},L_{1})}ℓ≡min(ℓ0,ℓ1){\displaystyle \ell \equiv \min(\ell _{0},\ell _{1})}

Then the series will:

Ali'smth ratio test

[edit]

This test is a direct extension of the second ratio test.[7][9] For0≤k≤m−1,{\displaystyle 0\leq k\leq m-1,} and positivean{\displaystyle a_{n}} define:

Lk≡limn→∞amn+kan{\displaystyle L_{k}\equiv \lim _{n\rightarrow \infty }{\frac {a_{mn+k}}{a_{n}}}}
L≡max(L0,L1,…,Lm−1){\displaystyle L\equiv \max(L_{0},L_{1},\ldots ,L_{m-1})}

By them{\displaystyle m}th ratio test, the series will:

If the above limits do not exist, it may be possible to use the limits superior and inferior. For0≤k≤m−1{\displaystyle 0\leq k\leq m-1} define:

Lk≡lim supn→∞amn+kan{\displaystyle L_{k}\equiv \limsup _{n\rightarrow \infty }{\frac {a_{mn+k}}{a_{n}}}}
ℓk≡lim infn→∞amn+kan{\displaystyle \ell _{k}\equiv \liminf _{n\rightarrow \infty }{\frac {a_{mn+k}}{a_{n}}}}
L≡max(L0,L1,…,Lm−1){\displaystyle L\equiv \max(L_{0},L_{1},\ldots ,L_{m-1})}ℓ≡min(ℓ0,ℓ1,…,ℓm−1){\displaystyle \ell \equiv \min(\ell _{0},\ell _{1},\ldots ,\ell _{m-1})}

Then the series will:

Ali--Deutsche Cohen φ-ratio test

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This test is an extension of them{\displaystyle m}th ratio test.[21]

Assume that the sequencean{\displaystyle a_{n}} is a positive decreasing sequence.

Letφ:Z+→Z+{\displaystyle \varphi :\mathbb {Z} ^{+}\to \mathbb {Z} ^{+}} be such thatlimn→∞nφ(n){\displaystyle \lim _{n\to \infty }{\frac {n}{\varphi (n)}}} exists. Denoteα=limn→∞nφ(n){\displaystyle \alpha =\lim _{n\to \infty }{\frac {n}{\varphi (n)}}}, and assume0<α<1{\displaystyle 0<\alpha <1}.

Assume also thatlimn→∞aφ(n)an=L.{\displaystyle \lim _{n\to \infty }{\frac {a_{\varphi (n)}}{a_{n}}}=L.}

Then the series will:

See also

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Footnotes

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  1. ^Weisstein, Eric W."Ratio Test".MathWorld.
  2. ^Rudin 1976, §3.34
  3. ^Apostol 1974, §8.14
  4. ^abcdefghBromwich, T. J. I'A (1908).An Introduction To The Theory of Infinite Series. Merchant Books.
  5. ^abcKnopp, Konrad (1954).Theory and Application of Infinite Series. London: Blackie & Son Ltd.
  6. ^abcTong, Jingcheng (May 1994). "Kummer's Test Gives Characterizations for Convergence or Divergence of all Positive Series".The American Mathematical Monthly.101 (5):450–452.doi:10.2307/2974907.JSTOR 2974907.
  7. ^abcdefAli, Sayel A. (2008)."The mth Ratio Test: New Convergence Test for Series".The American Mathematical Monthly.115 (6):514–524.doi:10.1080/00029890.2008.11920558.S2CID 16336333. Retrieved4 September 2024.
  8. ^abSamelson, Hans (November 1995). "More on Kummer's Test".The American Mathematical Monthly.102 (9):817–818.doi:10.2307/2974510.JSTOR 2974510.
  9. ^abcdefghBlackburn, Kyle (4 May 2012)."The mth Ratio Convergence Test and Other Unconventional Convergence Tests"(PDF). University of Washington College of Arts and Sciences. Retrieved27 November 2018.
  10. ^abcdefĎuriš, František (2009).Infinite series: Convergence tests (Bachelor's thesis). Katedra Informatiky, Fakulta Matematiky, Fyziky a Informatiky, Univerzita Komenského, Bratislava. Retrieved28 November 2018.
  11. ^abcĎuriš, František (2 February 2018). "On Kummer's test of convergence and its relation to basic comparison tests".arXiv:1612.05167 [math.HO].
  12. ^Weisstein, Eric W."Raabe's Test".MathWorld.
  13. ^Weisstein, Eric W."Bertrand's Test".MathWorld.
  14. ^Martin, Margaret (1941)."A sequence of limit tests for the convergence of series"(PDF).Bulletin of the American Mathematical Society.47 (6):452–457.doi:10.1090/S0002-9904-1941-07477-X.
  15. ^Abramov, Vyacheslav M. (May 2020). "Extension of the Bertrand–De Morgan test and its application".The American Mathematical Monthly.127 (5):444–448.arXiv:1901.05843.doi:10.1080/00029890.2020.1722551.S2CID 199552015.
  16. ^Weisstein, Eric W."Kummer's Test".MathWorld.
  17. ^Abramov, Vyacheslav, M. (21 June 2021). "A simple proof of Tong's theorem".arXiv:2106.13808 [math.HO].{{cite arXiv}}: CS1 maint: multiple names: authors list (link)
  18. ^Abramov, Vyacheslav M. (May 2022)."Evaluating the sum of convergent positive series"(PDF).Publications de l'Institut Mathématique. Nouvelle Série.111 (125):41–53.doi:10.2298/PIM2225041A.S2CID 237499616.
  19. ^Frink, Orrin (October 1948)."A ratio test".Bulletin of the American Mathematical Society.54 (10): 953.doi:10.1090/S0002-9904-1948-09111-X.
  20. ^Stark, Marceli (1949). "On the ratio test of Frink".Colloquium Mathematicum.2 (1):46–47.doi:10.4064/cm-2-1-46-47.
  21. ^Ali, Sayel; Cohen, Marion Deutsche (2012)."phi-ratio tests".Elemente der Mathematik.67 (4):164–168.doi:10.4171/EM/206.

References

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Precalculus
Limits
Differential calculus
Integral calculus
Vector calculus
Multivariable calculus
Sequences and series
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and numbers
History of calculus
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Miscellaneous topics
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