Calculus

Infinite Sequences and Series

Sequences and Series Logo

Alternating Series

A series in which successive terms have opposite signs is called an alternating series.

The Alternating Series Test (Leibniz's Theorem)

This test is the sufficient convergence test. It's also known as the Leibniz's Theorem for alternating series.

Let {an} be a sequence of positive numbers such that

  1. an+1 < an for all n;
  2. \(\lim\limits_{n \to \infty } {a_n} = 0.\)

Then the alternating series \(\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{a_n}} \) and \(\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}{a_n}} \) both converge.

Absolute and Conditional Convergence

A series \(\sum\limits_{n = 1}^\infty {{a_n}}\) is absolutely convergent, if the series \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) is convergent.

If the series \(\sum\limits_{n = 1}^\infty {{a_n}}\) is absolutely convergent then it is (just) convergent. The converse of this statement is false.

A series \(\sum\limits_{n = 1}^\infty {{a_n}}\) is called conditionally convergent, if the series is convergent but is not absolutely convergent.

Solved Problems

Example 1.

Use the alternating series test to determine the convergence of the series \[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n} \frac{{{{\sin }^2}n}}{n}}.\]

Solution.

By the alternating series test we find that

\[\lim\limits_{n \to \infty } \left| {{a_n}} \right| = \lim\limits_{n \to \infty } \left| {{{\left( { - 1} \right)}^n}\frac{{{{\sin }^2}n}}{n}} \right| = \lim\limits_{n \to \infty } \frac{{{{\sin }^2}n}}{n} = 0,\]

since \({\sin ^2}n \le 1.\) Hence, the given series converges.

Example 2.

Determine whether the series \[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{2n + 1}}{{3n + 2}}} \] is absolutely convergent, conditionally convergent, or divergent.

Solution.

We try to apply the alternating series test here:

\[\lim\limits_{n \to \infty } \left| {{a_n}} \right| = \lim\limits_{n \to \infty } \frac{{2n + 1}}{{3n + 2}} = \lim\limits_{n \to \infty } \frac{{\frac{{2n + 1}}{n}}}{{\frac{{3n + 2}}{n}}} = \lim\limits_{n \to \infty } \frac{{2 + \frac{1}{n}}}{{3 + \frac{2}{n}}} = \frac{2}{3} \ne 0.\]

The \(n\)th term does not approach \(0\) as \(n \to \infty.\) Therefore the given series diverges.

See more problems on Page 2.

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