Convergence tests

In mathematics, convergence tests are methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence or divergence of an infinite series.

1 List of tests
2 Comparison
3 Examples
4 Convergence of products
5 See also
6 References
7 External links

List of tests

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 (D'Alembert's criterion). Suppose that there exists $r$ such that

$\lim_{n \to \infty} \left|\frac{a_{n%2B1}}{a_n}\right| = r.$

If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.

Root test or nth root test (Cauchy's criterion). Define r as follows:

$r = \limsup_{n \to \infty}\sqrt[n]{|a_n|},$

where "lim sup" denotes the limit superior (possibly ∞; 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.

Integral test. The series can be compared to an integral to establish convergence or divergence. Let $f:[1,\infty)\to\R_%2B$ be a positive and monotone 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.

In other words, the series ${a_n}$ converges if and only if the integral converges.

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.

Limit comparison test. If $\left \{ a_n \right \}, \left \{ b_n \right \} > 0$ , and the limit $\lim_{n \to \infty} \frac{a_n}{b_n}$ exists and is not zero, then $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty b_n$ converges.

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

Alternating series test (Leibniz criterion)

For some specific types of series there are more specialized convergence tests, for instance for Fourier series there is the Dini test.

Comparison

The root test is stronger than the ratio test (it is more powerful because the required condition is weaker): whenever the ratio test determines the convergence or divergence of an infinite series, the root test does too, but not conversely.^[1]

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.

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 $\alpha > 1$ .

Convergence of products

While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. This can be achieved using following theorem: Let $\left \{ a_n \right \}_{n=1}^\infty$ be a sequence of positive numbers. Then the infinite product $\prod_{n=1}^\infty (1 %2B a_n)$ converges if and only if the series $\sum_{n=1}^\infty a_n$ converges. Also similarly, if $0 < a_n < 1$ holds, then $\prod_{n=1}^\infty (1 - a_n)$ approaches a non-zero limit if and only if the series $\sum_{n=1}^\infty a_n$ converges .

This can be proved by taking logarithm of the product and using limit comparison test.^[2]

Convergence tests

Contents

List of tests

Comparison

Examples

Convergence of products

See also

References

External links