Alternating Series Test

Alternating Series

Some series have alternating positive and negative terms. These are called alternating series. They are identified by the term ( – 1)n or ( – 1)n – 1 multiplying the terms an of the series. For the first type, with ( – 1)n, every even value of n makes the term positive whereas the odd powers of n make the term negative. In the 2nd type with ( – 1)n – 1, the opposite is true.

The Alternating Series Test

If the terms are decreasing -- so that for every k, we need only show that the limit of the nth term at infinity is zero to prove convergence. Any limit other than 0 means it diverges.

The Alternating Series Test (AST)

The series , with an > 0, converges if for every k, and Note: this test applies only to an alternating series. Before we apply the test, we must identify the series as alternating and we must state that the terms of the series are decreasing. Then we apply the test to the positive term series omitting the ( – 1) term.

Reminder: for limit at infinity of a fraction, divide through by the highest power of the variable, then set it = . Anything divided by infinity = 0.

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Examples:

Test these alternating series for convergence.

 a) it's easy to see that the denominator grows faster than the numerator so the terms are decreasing.Now we take the limit at infinity. , so series converges. b) again we can show the terms are decreasing, so we take the limit at infinity. , so series diverges.

Once we've shown the series (without the alternating sign) is decreasing, we can use any of the limiting techniques to evaluate the limit. Sometimes we have to use l'Hopital's Rule.

For more on this topic, see the lesson on Absolute and Conditional Convergence. Practice:

 1) 2) 3) Note: in #2, the ( – 1)n is hidden in the ( – 10)n and can be factored out to read ( – 1)n × 10. Solutions:

In all 3 cases, it's easy to show that the terms are decreasing so we need only find limit at .

1) since the denominator approaches . Series converges by AST.

2) so we take limit at for which = 0. Series converges.

3) so we take limit at for which = 0. Series converges. ( calculus 2 index )

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