Convergence tests

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Ratio test. Assume that for all n, $a n > 0$ . 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 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. Define r as follows:

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

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 $f (n) = a n$ be a positive and monotone decreasing function. 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.

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.

Abel's test

Raabe's test

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

1 Comparison
2 The Tests: When to use & Examples
3 Examples
4 References

[edit] 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 + ...

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

[edit] The Tests: When to use & Examples

http://www.math.cornell.edu/~alozano/calculus/testconvergence.pdf

[edit] Examples

Consider the series

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

Cauchy condensation test implies that (*) finitely convergent if

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

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 $2 (1 - α)$ . (**) is finitely convergent if its ratio is less than one (namely $α > 1$ ). Thus, (*) is finitely convergent if and only if $α > 1$ .