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Ratio test
In mathematical analysis, the ratio test is a test (or "criterion") for the convergence of a serieswhere each term is a real or complex number and all but finitely many terms are non-zero.
Ratio test
In mathematical analysis, the ratio test is a test (or "criterion") for the convergence of a series
\(\sum_{n=1}^\infty a_n,\)
where each term is a real or complex number and all but finitely many terms are non-zero. 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 limit
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
\(R = \lim\sup \left|\frac{a_{n+1}}{a_n}\right|\)
\(r = \lim\inf \left|\frac{a_{n+1}}{a_n}\right|\).
Then the ratio test states that:
- if R < 1, the series converges absolutely;
- if r > 1, the series diverges; or equivalently if \(\left|\frac{a_{n+1}}{a_n}\right|> 1\) 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 an does not approach zero;
- the test is otherwise inconclusive.
If the limit L in (1) exists, we must have L = R = r. So the original ratio test is a weaker version of the refined one.
Convergent because L < 1
Consider the series
\(\sum_{n=1}^\infty\frac{n}{e^n}\)
Applying the ratio test, one computes the limit
\(L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right| = \frac{1}{e} < 1.\)
Since this limit is less than 1, the series converges.
Divergent because L > 1
Consider the series
\(\sum_{n=1}^\infty\frac{e^n}{n}.\)
Putting this into the ratio test:
\(L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}} \right| = e > 1.\)
Thus the series diverges.
Inconclusive because L = 1
Consider the three series
\(\sum_{n=1}^\infty 1,\)
\(\sum_{n=1}^\infty \frac{1}{n^2},\)
\(\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}.\)
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 \(\left|\frac{a_{n+1}}{a_n}\right|\) of the three series are \(1,\) \(\frac{n^2}{(n+1)^2}\) and \(\frac{n}{n+1}\). So, in all three, the limit \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\) 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 generalized ratio test.
Suppose that \(r=\liminf_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|>1\). We also suppose that \((a_n)\) has infinite non-zero members, otherwise the series is just a finite sum hence it converges. Then there exists some \(\ell\in(1;r)\) such that there exists a natural number \(n_0\ge2\) satisfying \(a_{n_0}\ne0\) and \(\left|\frac{a_{n+1}}{a_n}\right|>\ell\) for all \(n\ge n_0\), because if no such \(\ell\) exists then there exists arbitrarily large \(n\) satisfying \(\left|\frac{a_{n+1}}{a_n}\right|<\ell\) for every \(\ell\in(1;r)\), then we can find a subsequence \(\left(a_{n_k}\right)_{k=1}^\infty\) satisfying \(\limsup_{n\to\infty}\left|\frac{a_{n_k+1}}{a_{n_k}} \right|\le\ell
Now suppose \(R=\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1\). Similar to the above case, we may find a natural number \(n_1\) and a \(c\in(R;1)\) such that \(|a_n|\le c^{n-n_1}\left|a_{n_1}\right|\) for \(n\ge n_1\). Then
\[\sum_{n=1}^\infty |a_n|=\sum_{k=1}^{n_1-1}|a_k|+\sum_{n=n_1}^\infty |a_n|\le\sum_{k=1}^{n_1-1}|a_k|+\sum_{n=n_1}^\infty c^{n-n_1}|a_{n_1}|=\sum_{k=1}^{n_1-1}|a_k|+\left|a_{n_1}\right|\sum_{n=0}^\infty c^n.\]
The series \(\sum_{n=0}^\infty c^n\) is the geometric series with common ratio \(c\in(0;1)\), hence \(\sum_{n=0}^\infty c^n=\frac{1}{1-c}\) which is finite. The sum \(\sum_{k=1}^{n_1-1}|a_k|\) is a finite sum and hence it is bounded, this implies the series \(\sum_{n=1}^\infty |a_n|\) converges by the monotone convergence theorem and the series \(\sum_{n=1}^\infty a_n\) converges by the absolute convergence test.
When the limit \(\left|\frac{a_{n+1}}{a_n}\right|\) exists and equals to \(L\) then \(r=R=L\), 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 \(\sum a_n\) 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:
\(\sum_{n=1}^\infty a_n = \sum_{n=1}^N a_n+\sum_{n=N+1}^\infty a_n\)
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 (\(\rho_n\)) 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 \(\lim_{n\to\infty}\rho_n\).
All of the tests have regions in which they fail to describe the convergence properties of \(\sum a_n\). In fact, no convergence test can fully describe the convergence properties of the series. This is because if \(\sum a_n\) is convergent, a second convergent series \(\sum b_n\) can be found which converges more slowly: i.e., it has the property that \(\lim_{n\to\infty}\frac{b_n}{a_n}=\infty\). Furthermore, if \(\sum a_n\) is divergent, a second divergent series \(\sum b_n\) can be found which diverges more slowly: i.e., it has the property that \(\lim_{n\to\infty}\frac{b_n}{a_n}=0\). 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
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.
Tong's modification of Kummer's test
A new version of Kummer's test was established by Tong. See also 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 \(\sum_{n=1}^\infty a_n\) converges if and only if there exists a positive sequence \(\zeta_n\), \(n=1,2,\dots\), such that \(\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}\geq c>0.\)
- Series \(\sum_{n=1}^\infty a_n\) diverges if and only if there exists a positive sequence \(\zeta_n\), \(n=1,2,\dots\), such that \(\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}\leq0,\) and \(\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty.\)
The first of these statements can be simplified as follows:
- Series \(\sum_{n=1}^\infty a_n\) converges if and only if there exists a positive sequence \(\zeta_n\), \(n=1,2,\dots\), such that \(\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}=1.\)
The second statement can be simplified similarly:
- Series \(\sum_{n=1}^\infty a_n\) diverges if and only if there exists a positive sequence \(\zeta_n\), \(n=1,2,\dots\), such that \(\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}=0,\) and \(\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty.\)
However, it becomes useless, since the condition \(\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty\) in this case reduces to the original claim \(\sum_{n=1}^{\infty}a_n=\infty.\)
Frink's ratio test
Another ratio test that can be set in the framework of Kummer's theorem was presented by Orrin Frink 1948.
Suppose \(a_n\) is a sequence in \(\mathbb{C}\setminus\{0\}\),
- If \(\limsup_{n\rightarrow\infty}\Big(\frac{|a_{n+1}|}{|a_n|}\Big)^n<\frac1e\), then the series \(\sum_na_n\) converges absolutely.
- If there is \(N\in\mathbb{N}\) such that \(\Big(\frac{|a_{n+1}|}{|a_n|}\Big)^n\geq\frac1e\) 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.
Ali's second ratio test
A more refined ratio test is the second ratio test: For \(a_n>0\) define:
By the second ratio test, the series will:
- Converge if \(L<\frac{1}{2}\)
- Diverge if \(L>\frac{1}{2}\)
- If \(L=\frac{1}{2}\) then the test is inconclusive.
If the above limits do not exist, it may be possible to use the limits superior and inferior. Define:
Then the series will:
- Converge if \(L<\frac{1}{2}\)
- Diverge if \(\ell>\frac{1}{2}\)
- If \(\ell \le \frac{1}{2} \le L\) then the test is inconclusive.
Ali's mth ratio test
This test is a direct extension of the second ratio test. For \(0\leq k\leq m-1,\) and positive \(a_n\) define:
By the \(m\)th ratio test, the series will:
- Converge if \(L<\frac{1}{m}\)
- Diverge if \(L>\frac{1}{m}\)
- If \(L=\frac{1}{m}\) 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:
Then the series will:
- Converge if \(L<\frac{1}{m}\)
- Diverge if \(\ell>\frac{1}{m}\)
- If \(\ell \leq \frac{1}{m} \leq L\), then the test is inconclusive.
Ali--Deutsche Cohen φ-ratio test
This test is an extension of the \(m\)th ratio test.
Assume that the sequence \(a_n\) is a positive decreasing sequence.
Let \(\varphi:\mathbb{Z}^+\to\mathbb{Z}^+\) be such that \(\lim_{n\to\infty}\frac{n}{\varphi(n)}\) exists. Denote \(\alpha=\lim_{n\to\infty}\frac{n}{\varphi(n)}\), and assume \(0<\alpha<1\).
Assume also that \(\lim_{n\to\infty}\frac{a_{\varphi(n)}}{a_n}=L.\)
Then the series will:
- Converge if \(L<\alpha\)
- Diverge if \(L>\alpha\)
- If \(L=\alpha\), then the test is inconclusive.
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Questions posées par les gens
What is the ε, δ definition actually saying?
That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.
Why does the harmonic series diverge when its terms go to zero?
Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.
Les parties de cette page sont adaptées à partir Wikipedia (CC BY-SA 4.0). Condensés et réexpliqués ici ; les erreurs sont à nous.
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