Alternating Series

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The Alternating Series Test

Suppose that a weight from a spring is released. Let a₁ be the distance that the spring drops on the first bounce. Let a₂ be the amount the weight travels up the first time. Let a₃ be the amount the weight travels on the way down for the second trip. Let a₄ be the amount that the weight travels on the way up for the second trip, etc.

$Picture of the line segments a1, a2, a3, a4, a5, a6, ... The a1 bottom matches a2 bottom, a2 top matches a3 top, a3 bottom matches a4 bottom, a4 top matches a5 top, a5 bottom matches a6 bottom.$

Then eventually the weight will come to rest somewhere in the middle. This leads us to

Theorem: The Alternating Series Test

Let a_n > 0 for all n and suppose that the following two conditions hold:

{ a_n } is a decreasing sequence for large n.
$ \lim\limits_{n\to \infty} a_n = 0 $

Then the corresponding series $ \displaystyle\sum_{n=1}^{\infty}(-1)^n a_n $ and $\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1} a_n $

converge.

Proof:

We will prove the theorem for the second given series. This is enough, since the first can be obtained from the second just by multiplying by -1. We look at the series as adding two at a time and then adding them all together.

        s_2n = (a₁ - a₂) + (a₃ - a₄) + ...+ (a_2n_-1 - a_2n) > 0

which shows that this is bounded below by 0. Now single out the first term and then add the rest two at a time

        s_2n = a₁ - (a₂ - a₃) - (a₄ - a₅) - ...- (a_2n_-2 - a_2n_-1) - a_2n

= a₁ - [(a₂ - a₃) + (a₄ - a₅) + ...+ (a_2n_-2 - a_2n_-1) + a_2n]

This second equation subtracts a positive number from the first term. Hence

        s_2n < a₁

which shows that the sequence is bounded above by a₁. Notice that s_2n is monotonic since each difference is positive. Therefore s_2n  is bounded and monotonic and thus converges. Since the a_n tend toward zero as n tends towards infinity, we have



The limit of the partial sums exist and hence the series converges.

Example

        $ \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n} $
converges by the alternating series test, since the

          $ \lim\limits_{n\to \infty} \frac{1}{n} = 0 $

and

           1                1
                    >
           n              n + 1

Exercises:

Determine whether the following converge:

$ \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!} $
$ \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n n^2}{n+1} $

The Remainder Theorem

Consider the spring example again. The weight will always be between the two previous positions. Hence we have

      The Remainder Theorem

Let

          $ \displaystyle\sum_{n=1}^{\infty}(-1)^n a_n $

then

          |L - s_n|  < a_{n + 1}

This says that the error in using n terms to approximate an alternating series is always less then the n + 1^st term.

Example

Use a calculator to determine

$ \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} $

With an error of less than .01.

Solution:

We have

         Error < .01

so choose n such that

             1
                   < .01
            n

Here, n = 101 will work. Then use your calculator to get  0.70.

Absolute and Conditional Convergence

Definitions

$ \displaystyle\sum_{n=1}^{\infty}(-1)^{n}a_n $ is called absolutely convergent if $ \displaystyle\sum_{n=1}^{\infty}a_n $ converges.
$ \displaystyle\sum_{n=1}^{\infty}(-1)^{n}a_n $ is called conditionally convergent if $ \displaystyle\sum_{n=1}^{\infty}(-1)^{n}a_n $
converges, but $ \displaystyle\sum_{n=1}^{\infty}a_n $ diverges.

Example:

The alternating harmonic series is conditionally convergent

        $ \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} $

since we saw before that it converges by the alternating series test but its absolute value (the harmonic series) diverges.

The series

$ \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^2} $

is absolutely convergent since the series of the absolute value of its terms is a P-series with p = 2, hence converges.

The Rearrangement Theorem

Let $ \displaystyle\sum_{n=1}^{\infty}(-1)^n a_n $ be a conditionally convergent series and let k be a real number. Then there exists a rearrangement of the terms so that you add them up and end up with k. As strange as it may seem, addition is not commutative for conditionally convergent series. On the other hand for absolutely convergent series any rearrangement produces the same limit.

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