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2006-08-17 04:13:26 · 6 answers · asked by loque . 1 in Science & Mathematics Mathematics

6 answers

Yup. They're used to determine if a particular series (or sequence) of numbers, polynomials, functions, etc. converges to some finite value in the limit.


Doug

2006-08-17 04:18:46 · answer #1 · answered by doug_donaghue 7 · 0 0

A relatively simple test that I learned in school is to look at the ratio of two adjacent terms of the series. If this ratio is less than 1, the series converges.

For example, you know that the series 2^-i sums to 2 where i ranges from 0 to infinity.

The ratio between term k and term k+1 will always be 1/2, and this shows that the series converges.

Mathworld has a link with a bit more detail and rigor.

2006-08-17 14:59:34 · answer #2 · answered by arbeit 4 · 0 0

To test divergence, you usually have to write out the first few terms of a series, and then compile an 'nth' term. The n'th term has to satisfy some requirement for it be divergent or convergant, I forget what that requirement was.

Convergant: the series adds to a specific value. Kind of like taking the integral of a function and finding that it has a limit.

Divergant; the series adds to infinity.

I hope I'm right. Last time I did seq/series was 5 years ago.

2006-08-17 11:37:19 · answer #3 · answered by Krzysztof_98 2 · 0 0

One of the tests to determine if a series diverges is the comparison test. This is where you observe the the absolute value of the first few terms of the series in question are greater than the absolute value of corresponding terms in a series that you know diverges. Then you show that the absolute value of the n th term in the series in question is greater than the absolute value of the n th term in the series that is known to diverge.

The term 'absolute value' is used since series can diverge to negative infinity.

2006-08-17 12:40:17 · answer #4 · answered by williamh772 5 · 0 0

Usually the ratio test determines whether a series diverges if the sum of (n+1) terms divided by the sum of (n) terms is greater than or equal to 1.

2006-08-21 10:32:35 · answer #5 · answered by Anonymous · 0 0

http://mathworld.wolfram.com/ConvergenceTests.html

2006-08-17 11:19:34 · answer #6 · answered by Bruno 3 · 0 0

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