User:Editeur24/convergence testexamples

In mathematics, convergence tests are methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence or divergence of an infinite series $$\sum_{n=1}^\infty a_n$$.

Limit of the summand
If the limit of the summand is undefined or nonzero, that is $$\lim_{n \to \infty}a_n \ne 0$$, then the series must diverge. In this sense, the partial sums are Cauchy only if this limit exists and is equal to zero. The test is inconclusive if the limit of the summand is zero.

Ratio test
This is also known as d'Alembert's criterion.


 * Suppose that there exists $$r$$ such that
 * $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = r.$$
 * If r < 1, then the series is absolutely convergent. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge.
 * The power series 1, .5, .25, .125,... has $$r = |.5^{n+1}/.5^n| = .5,$$ and hence converges. If the series were $$2^n,$$ it would diverge. If it were $$1^n,$$ the ratio test would fail to give an answer (though obviously that series does diverge).

Root test
This is also known as the nth root test or Cauchy's criterion.


 * Let
 * $$r=\limsup_{n\to\infty}\sqrt[n]{|a_n|},$$
 * where $$\limsup$$ denotes the limit superior (possibly $$\infty$$; if the limit exists it is the same value).
 * If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the root test is inconclusive, and the series may converge or diverge.

The root test is stronger than the ratio test: whenever the ratio test determines the convergence or divergence of an infinite series, the root test does too, but not conversely.

For example, for the series


 * 1 + 1 + 0.5 + 0.5 + 0.25 + 0.25 + 0.125 + 0.125 + ... = 4

convergence follows from the root test but not from the ratio test.

Integral test
The series can be compared to an integral to establish convergence or divergence. Let $$f:[1,\infty)\to\R_+$$ be a non-negative and monotonically decreasing function such that $$f(n) = a_n$$.
 * If
 * $$\int_1^\infty f(x) \, dx=\lim_{t\to\infty}\int_1^t f(x) \, dx<\infty,$$
 * then the series converges. But if the integral diverges, then the series does so as well. Thus, the series $${a_n}$$ converges if and only if the integral converges.
 * The harmonic series 1, 1/2, 1/3, 1/4,... has $$\int_1^\infty (1/x) dx = |_1^\infty  log(x) = \infty$$ and so does not converge. The integral test succeeds here where the ratio test fails, because $$r =  lim_{n \rightarrow \infty} | (1/(n+1))/(1/n)|  = lim_{n \rightarrow \infty}  n/(n+1) =1$$. The root test also fails, because $$r =  lim_{n \rightarrow \infty}  \sqrt[n]{|a_n|}  = lim_{n \rightarrow \infty} \sqrt[n]{1/n} =1.$$

Direct comparison test
If the series $$\sum_{n=1}^\infty b_n$$ is an absolutely convergent series and $$|a_n|\le |b_n|$$ for sufficiently large n, then the series $$\sum_{n=1}^\infty a_n$$ converges absolutely.

The harmonic series $$a_n =$$ 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8,... can be compared to the series $$b_n =$$  1/2, 1/4, 1/4, 1/8, 1/8, 1/8, 1/8, 1/16,...,  where $$a_n>b_n$$ for any $$n.$$ The sum of the $$b_n$$ is infinite because it is 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8)+ (1/16 +..., an infinite sum of 1/2's, so the sum of the bigger $$a_n$$ must also be infinite. As noted above, the ratio test and root test fail for the harmonic series.

Limit comparison test
If $$\{a_n\},\{b_n\}>0$$, (that is, each element of the two sequences is positive) and the limit $$\lim_{n\to\infty} \frac{a_n}{b_n}$$ exists, is finite and non-zero, then $$\sum_{n=1}^\infty a_n$$ diverges if and only if $$\sum_{n=1}^\infty b_n$$ diverges. '''

Cauchy condensation test
Let $$\left \{ a_n \right \}$$ be a positive non-increasing sequence. Then the sum $$A = \sum_{n=1}^\infty a_n$$ converges if and only if the sum $$A^* = \sum_{n=0}^\infty 2^n a_{2^n}$$ converges. Moreover, if they converge, then $$A \leq A^* \leq 2A$$ holds.

Abel's test
Suppose the following statements are true:


 * 1) $$\sum a_n $$ is a convergent series,
 * 2) $$\left\{b_n\right\}$$ is a monotonic sequence, and
 * 3) $$\left\{b_n\right\}$$ is bounded.

Then $$\sum a_nb_n $$ is also convergent.

Absolute convergence test
Every absolutely convergent series converges.

Alternating series test
This is also known as the Leibniz criterion.

Suppose the following statements are true:


 * 1) $$ \lim_{n \to \infty} a_n = 0 $$,
 * 2) for every n, $$ a_{n+1} \le a_n $$

Then $$ \sum_{n = k}^\infty (-1)^{n} a_n $$ and $$ \sum_{n = k}^\infty (-1)^{n+1} a_n $$ are convergent series.

Dirichlet's test
If $$\{a_n\}$$ is a sequence of real numbers and $$\{b_n\}$$ a sequence of complex numbers satisfying


 * $$a_n \geq a_{n+1}$$


 * $$\lim_{n \rightarrow \infty}a_n = 0$$


 * $$\left|\sum^{N}_{n=1}b_n\right|\leq M$$ for every positive integer N

where M is some constant, then the series


 * $$\sum^{\infty}_{n=1}a_n b_n$$

converges.

Raabe–Duhamel's test
Let $$a_n>0$$.

Define


 * $$b_n=n\left(\frac{a_n}{a_{n+1}}-1 \right).$$

If


 * $$L=\lim_{n\to\infty}b_n$$

exists there are three possibilities:


 * if L > 1 the series converges
 * if L < 1 the series diverges
 * and if L = 1 the test is inconclusive.

An alternative formulation of this test is as follows. Let { an } be a series of real numbers. Then if b > 1 and K (a natural number) exist such that


 * $$\left|\frac{a_{n+1}}{a_n}\right|\le 1-\frac{b}{n} $$

for all n > K then the series {an} is convergent.

Bertrand's test
Let { an } be a sequence of positive numbers.

Define


 * $$b_n=\ln n\left(n\left(\frac{a_n}{a_{n+1}}-1 \right)-1\right).$$

If


 * $$L=\lim_{n\to\infty}b_n$$

exists, there are three possibilities:


 * if L > 1 the series converges
 * if L < 1 the series diverges
 * and if L = 1 the test is inconclusive.

Gauss's test
Let { an } be a sequence of positive numbers. If $$\frac{a_n}{a_{n + 1}} = 1+ \frac{\alpha}{n} + O(1/n^\beta)$$ for some β > 1, then $$ \sum a_n$$ converges if $α > 1$ and diverges if $α ≤ 1$.

Examples
Consider the series

$$(*) \;\;\; \sum_{n=1}^{\infty} \frac{1}{n^\alpha}.$$

Cauchy condensation test implies that (*) is finitely convergent if


 * $$ (**) \;\;\; \sum_{n=1}^\infty 2^n \left( \frac 1 {2^n}\right)^\alpha $$

is finitely convergent. Since


 * $$\sum_{n=1}^\infty 2^n \left( \frac 1 {2^n} \right)^\alpha = \sum_{n=1}^\infty 2^{n-n\alpha} = \sum_{n=1}^\infty 2^{(1-\alpha) n} $$

(**) is geometric series with ratio $$ 2^{(1-\alpha)} $$. (**) is finitely convergent if its ratio is less than one (namely $$\alpha > 1$$). Thus, (*) is finitely convergent if and only if $$ \alpha > 1 $$.