Ratio test - WikiHQ

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

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

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

thumb|Decision diagram for the ratio test

The usual form of the test makes use of the limit

{a_n}\right|.</math>|

The 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

:<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:

if R < 1, the series converges absolutely;
if r > 1, the series diverges; or equivalently if <math>\left|\frac{a_{n+1{a_n}\right|> 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 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

:<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 (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 <math>\left|\frac{a_{n+1{a_n}\right|</math> of the three series are <math>1,</math>   <math>\frac{n^2}{(n+1)^2}</math>    and   <math>\frac{n}{n+1}</math>. So, in all three, 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: the ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence.

Proof

thumb|300px|In this example, the ratio of adjacent terms in the blue sequence converges to L=1/2. We choose r = (L+1)/2 = 3/4. Then the blue sequence is dominated by the red sequence 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.

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.

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.

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 allow one to deal with this case.

In all the tests below one assumes that Σan is a sum with positive an. 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 aN 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 (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. the series will:

Converge if <math>\rho=\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. 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. Extended Bertrand's test

This extension probably appeared at the first time by Margaret Martin in 1941. 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.

Let <math>K\geq1</math> be an integer, and let <math>\ln_{(K)}(x)</math> denote the <math>K</math>th iterate of natural logarithm, i.e. <math>\ln_{(1)}(x)=\ln (x)</math> and for any <math>2\leq k\leq K</math>,

<math>\ln_{(k)}(x)=\ln_{(k-1)}(\ln (x))</math>.

Suppose that the ratio <math>a_n/a_{n+1}</math>, when <math>n</math> is large, can be presented in the form

:<math>\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\geq1.</math>

(The empty sum is assumed to be 0. With <math>K=1</math>, the test reduces to Bertrand's test.)

The value <math>\rho_{n}</math> can be presented explicitly in the form

:<math>\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).</math>

Extended Bertrand's test asserts that the series

Converge when there exists a <math>c>1</math> such that <math>\rho_n \geq c</math> for all <math>n>N</math>.
Diverge when <math>\rho_n \leq 1</math> for all <math>n>N</math>.
Otherwise, the test is inconclusive.

For the limit version, the series

Converge if <math>\rho=\lim_{n\to\infty}\rho_n>1</math> (this includes the case <math>\rho = \infty</math>)
Diverge if <math>\lim_{n\to\infty}\rho_n<1</math>.
If <math>\rho = 1</math>, 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 <math>\liminf \rho_n > 1</math>
Diverge if <math>\limsup \rho_n < 1</math>
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 an > 0 and r > 1, if a bounded sequence Cn can be found such that for all n:

for further discussions and new proofs. The provided modification of Kummer's theorem characterizes

all 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 <math>\sum_{n=1}^\infty a_n</math> converges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1-\zeta_{n+1}\geq c>0.</math>

Series <math>\sum_{n=1}^\infty a_n</math> diverges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1-\zeta_{n+1}\leq0,</math> and <math>\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty.</math>

The first of these statements can be simplified as follows:

Series <math>\sum_{n=1}^\infty a_n</math> converges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1-\zeta_{n+1}=1.</math>

The second statement can be simplified similarly:

Series <math>\sum_{n=1}^\infty a_n</math> diverges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1-\zeta_{n+1}=0,</math> and <math>\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty.</math>

However, it becomes useless, since the condition <math>\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty</math> in this case reduces to the original claim <math>\sum_{n=1}^{\infty}a_n=\infty.</math>

Frink's ratio test

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

Suppose <math>a_n</math> is a sequence in <math>\mathbb{C}\setminus\{0\}</math>,

If <math> \limsup_{n\rightarrow\infty}\Big(\frac{|a_{n+1}|}{|a_n|}\Big)^n<\frac1e </math>, then the series <math>\sum_na_n</math> converges absolutely.
If there is <math>N\in\mathbb{N}</math> such that <math> \Big(\frac{|a_{n+1}|}{|a_n|}\Big)^n\geq\frac1e </math> for all <math>n\geq N</math>, then <math>\sum_n|a_n|</math> diverges.

This result reduces to a comparison of <math>\sum_n|a_n|</math> with a power series <math>\sum_n n^{-p}</math>, and can be seen to be related to Raabe's test.

Ali's second ratio test

A more refined ratio test is the second ratio test:

Assume that the sequence <math>a_n</math> is a positive decreasing sequence.

Let <math>\varphi:\mathbb{Z}^+\to\mathbb{Z}^+</math> be such that <math>\lim_{n\to\infty}\frac{n}{\varphi(n)}</math> exists. Denote <math>\alpha=\lim_{n\to\infty}\frac{n}{\varphi(n)}</math>, and assume <math>0<\alpha<1</math>.

Assume also that <math>\lim_{n\to\infty}\frac{a_{\varphi(n){a_n}=L.</math>

Then the series will:

Converge if <math>L<\alpha</math>
Diverge if <math>L>\alpha</math>
If <math>L=\alpha</math>, then the test is inconclusive.

Footnotes

References

.
: §8.14.
: §3.3, 5.4.
: §3.34.
: §2.36, 2.37.

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