Alternating Series

So far in this chapter, we have primarily discussed series with positive terms. In this section we introduce alternating series—those series whose terms alternate in sign. We will show in a later chapter that these series often arise when studying power series. After defining alternating series, we introduce the alternating series test to determine whether such a series converges.

The Alternating Series Test

A series whose terms alternate between positive and negative values is an alternating series. For example, the series

Series (1), shown in [link], is a geometric series. Since

\| r \| = \| − 1 / 2 \| < 1,

the series converges. Series (2), shown in [link], is called the alternating harmonic series. We will show that whereas the harmonic series diverges, the alternating harmonic series converges.

Proof

is a decreasing sequence that is bounded below, by the Monotone Convergence Theorem,

{S_{2 k + 1}}

Since the odd terms and the even terms in the sequence of partial sums converge to the same limit

S,

More generally, any alternating series of form (3) ([link]) or (4) ([link]) converges as long as

b_{1} \geq b_{2} \geq b_{3} \geq ⋯

([link]). The proof is similar to the proof for the alternating harmonic series.

We remark that this theorem is true more generally as long as there exists some integer

N

Remainder of an Alternating Series

It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series

be the corresponding sequence of partial sums. From [link], we see that for any integer

N \geq 1,

In other words, if the conditions of the alternating series test apply, then the error in approximating the infinite series by the

N th

Absolute and Conditional Convergence

Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} / n .

The series whose terms are the absolute value of these terms is the harmonic series, since

\sum_{n = 1}^{\infty} \| {(−1)}^{n + 1} / n \| = \sum_{n = 1}^{\infty} 1 / n .

Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.

By comparison, consider the series

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} / n^{2} .

The series whose terms are the absolute values of the terms of this series is the series

\sum_{n = 1}^{\infty} 1 / n^{2} .

Since both of these series converge, we say the series

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} / n^{2}

As shown by the alternating harmonic series, a series

\sum_{n = 1}^{\infty} a_{n}

may diverge. In the following theorem, however, we show that if

\sum_{n = 1}^{\infty} \| a_{n} \|

Proof

Consequently, by the comparison test, since

2 \sum_{n = 1}^{\infty} \| a_{n} \|

converges. By using the algebraic properties for convergent series, we conclude that

To see the difference between absolute and conditional convergence, look at what happens when we rearrange the terms of the alternating harmonic series

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} / n .

We show that we can rearrange the terms so that the new series diverges. Certainly if we rearrange the terms of a finite sum, the sum does not change. When we work with an infinite sum, however, interesting things can happen.

Begin by adding enough of the positive terms to produce a sum that is larger than some real number

M > 0 .

Then add more positive terms until the sum reaches 100. That is, find another integer

j > k

Continuing in this way, we have found a way of rearranging the terms in the alternating harmonic series so that the sequence of partial sums for the rearranged series is unbounded and therefore diverges.

The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In [link], we show how to rearrange the terms to create a new series that converges to

3 ln (2) / 2 .

We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number

r;

that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series

\sum_{n = 1}^{\infty} a_{n}

is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.

Key Concepts

Key Equations

State whether each of the following series converges absolutely, conditionally, or not at all.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{n}{n + 3}

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{\sqrt{n} + 1}{\sqrt{n} + 3}

Does not converge by divergence test. Terms do not tend to zero.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{1}{\sqrt{n + 3}}

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{\sqrt{n + 3}}{n}

Converges conditionally by alternating series test, since $\sqrt{n + 3} / n$

is decreasing. Does not converge absolutely by comparison with p-series, $p = 1 / 2 .$

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{1}{n!}

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{3^{n}}{n!}

Converges absolutely by limit comparison to $3^{n} / 4^{n},$

for example.

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

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {(\frac{n + 1}{n})}^{n}

Diverges by divergence test since $lim_{n \to \infty} \| a_{n} \| = e .$

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {sin}^{2} n

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {cos}^{2} n

Does not converge. Terms do not tend to zero.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {sin}^{2} (1 / n)

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {cos}^{2} (1 / n)

lim_{n \to \infty} {cos}^{2} (1 / n) = 1 .

Diverges by divergence test.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} ln (1 / n)

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} ln (1 + \frac{1}{n})

Converges by alternating series test.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{n^{2}}{1 + n^{4}}

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{n^{e}}{1 + n^{π}}

Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, $p = π - e$

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} 2^{1 / n}

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n^{1 / n}

Diverges; terms do not tend to zero.

\sum_{n = 1}^{\infty} {(−1)}^{n} (1 - n^{1 / n})

(Hint: $n^{1 / n} \approx 1 + ln (n) / n$

for large $n .)$

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n (1 - cos (\frac{1}{n}))

(Hint: $cos (1 / n) \approx 1 - 1 / n^{2}$

for large $n .)$

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} (\sqrt{n + 1} - \sqrt{n})

(Hint: Rationalize the numerator.)

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})

(Hint: Find common denominator then rationalize numerator.)

Converges absolutely by limit comparison with p-series, $p = 3 / 2,$

after applying the hint.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} (ln (n + 1) - ln n)

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n ({tan}^{−1} (n + 1) - {tan}^{−1} n)

(Hint: Use Mean Value Theorem.)

Converges by alternating series test since $n ({tan}^{−1} (n + 1) − {tan}^{−1} n)$

is decreasing to zero for large $n .$

Does not converge absolutely by limit comparison with harmonic series after applying hint.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} ({(n + 1)}^{2} - n^{2})

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

Converges absolutely, since $a_{n} = \frac{1}{n} - \frac{1}{n + 1}$

are terms of a telescoping series.

\sum_{n = 1}^{\infty} \frac{cos (n π)}{n}

\sum_{n = 1}^{\infty} \frac{cos (n π)}{n^{1 / n}}

Terms do not tend to zero. Series diverges by divergence test.

\sum_{n = 1}^{\infty} \frac{1}{n} sin (\frac{n π}{2})

\sum_{n = 1}^{\infty} sin (n π / 2) sin (1 / n)

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

In each of the following problems, use the estimate $\| R_{N} \| \leq b_{N + 1}$

to find a value of $N$

that guarantees that the sum of the first $N$

terms of the alternating series $\sum_{n = 1}^{\infty} {(−1)}^{n + 1} b_{n}$

differs from the infinite sum by at most the given error. Calculate the partial sum $S_{N}$

for this $N .$

[T] $b_{n} = 1 / n,$

error $< 10^{−5}$

[T] $b_{n} = 1 / ln (n),$

n \geq 2,

error $< 10^{−1}$

ln (N + 1) > 10,

N + 1 > e^{10},

N \geq 22026;

S_{22026} = 0.0257 …

[T] $b_{n} = 1 / \sqrt{n},$

error $< 10^{−3}$

[T] $b_{n} = 1 / 2^{n},$

error $< 10^{−6}$

2^{N + 1} > 10^{6}

or $N + 1 > 6 ln (10) / ln (2) = 19.93 .$

or $N \geq 19;$

S_{19} = 0.333333969 …

[T] $b_{n} = ln (1 + \frac{1}{n}),$

error $< 10^{−3}$

[T] $b_{n} = 1 / n^{2},$

error $< 10^{−6}$

{(N + 1)}^{2} > 10^{6}

or $N > 999;$

S_{1000} \approx 0.822466 .

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

If $b_{n} \geq 0$

is decreasing and $lim_{n \to \infty} b_{n} = 0,$

then $\sum_{n = 1}^{\infty} (b_{2 n - 1} - b_{2 n})$

converges absolutely.

If $b_{n} \geq 0$

is decreasing, then $\sum_{n = 1}^{\infty} (b_{2 n - 1} - b_{2 n})$

converges absolutely.

True. $b_{n}$

need not tend to zero since if $c_{n} = b_{n} - lim b_{n},$

then $c_{2 n - 1} - c_{2 n} = b_{2 n - 1} - b_{2 n} .$

If $b_{n} \geq 0$

and $lim_{n \to \infty} b_{n} = 0$

then $\sum_{n = 1}^{\infty} (\frac{1}{2} (b_{3 n - 2} + b_{3 n - 1}) − b_{3 n})$

converges.

If $b_{n} \geq 0$

is decreasing and $\sum_{n = 1}^{\infty} (b_{3 n - 2} + b_{3 n - 1} - b_{3 n})$

converges then $\sum_{n = 1}^{\infty} b_{3 n - 2}$

converges.

True. $b_{3 n - 1} - b_{3 n} \geq 0,$

so convergence of $\sum b_{3 n - 2}$

follows from the comparison test.

If $b_{n} \geq 0$

is decreasing and $\sum_{n = 1}^{\infty} {(−1)}^{n - 1} b_{n}$

converges conditionally but not absolutely, then $b_{n}$

does not tend to zero.

Let $a_{n}^{+} = a_{n}$

if $a_{n} \geq 0$

and $a_{n}^{-} = − a_{n}$

if $a_{n} < 0 .$

(Also, $a_{n}^{+} = 0 if a_{n} < 0$

and $a_{n}^{-} = 0 if a_{n} \geq 0 .)$

If $\sum_{n = 1}^{\infty} a_{n}$

converges conditionally but not absolutely, then neither $\sum_{n = 1}^{\infty} a_{n}^{+}$

nor $\sum_{n = 1}^{\infty} a_{n}^{-}$

converge.

True. If one converges, then so must the other, implying absolute convergence.

Suppose that $a_{n}$

is a sequence of positive real numbers and that $\sum_{n = 1}^{\infty} a_{n}$

converges.

Suppose that $b_{n}$

is an arbitrary sequence of ones and minus ones. Does $\sum_{n = 1}^{\infty} a_{n} b_{n}$

necessarily converge?

Suppose that $a_{n}$

is a sequence such that $\sum_{n = 1}^{\infty} a_{n} b_{n}$

converges for every possible sequence $b_{n}$

of zeros and ones. Does $\sum_{n = 1}^{\infty} a_{n}$

converge absolutely?

Yes. Take $b_{n} = 1$

if $a_{n} \geq 0$

and $b_{n} = 0$

if $a_{n} < 0 .$

Then $\sum_{n = 1}^{\infty} a_{n} b_{n} = \sum_{n : a_{n} \geq 0} a_{n}$

converges. Similarly, one can show $\sum_{n : a_{n} < 0} a_{n}$

converges. Since both series converge, the series must converge absolutely.

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{{sin}^{2} n}{n}

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{{cos}^{2} n}{n}

Not decreasing. Does not converge absolutely.

1 + \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \frac{1}{6} - \frac{1}{7} - \frac{1}{8} + ⋯

1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ⋯

Not alternating. Can be expressed as $\sum_{n = 1}^{\infty} (\frac{1}{3 n - 2} + \frac{1}{3 n - 1} - \frac{1}{3 n}),$

which diverges by comparison with $\sum \frac{1}{3 n - 2} .$

Show that the alternating series $1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{4} - \frac{1}{8} + ⋯$

does

not converge. What hypothesis of the alternating series test is not met?

Suppose that $\sum a_{n}$

converges absolutely. Show that the series consisting of the positive terms $a_{n}$

also converges.

Let $a^{+}_{n} = a_{n}$

if $a_{n} \geq 0$

and $a^{+}_{n} = 0$

if $a_{n} < 0 .$

Then $a^{+}_{n} \leq \| a_{n} \|$

for all $n$

so the sequence of partial sums of $a^{+}_{n}$

is increasing and bounded above by the sequence of partial sums of $\| a_{n} \|,$

which converges; hence, $\sum_{n = 1}^{\infty} a^{+}_{n}$

converges.

Show that the alternating series $\frac{2}{3} - \frac{3}{5} + \frac{4}{7} - \frac{5}{9} + ⋯$

does not converge. What hypothesis of the alternating series test is not met?

The formula $cos θ = 1 - \frac{θ^{2}}{2!} + \frac{θ^{4}}{4!} - \frac{θ^{6}}{6!} + ⋯$

will be derived in the next chapter. Use the remainder $\| R_{N} \| \leq b_{N + 1}$

to find a bound for the error in estimating $cos θ$

by the fifth partial sum $1 - θ^{2} / 2! + θ^{4} / 4! − θ^{6} / 6! + θ^{8} / 8!$

for $θ = 1,$

θ = π / 6,

and $θ = π .$

For $N = 5$

one has $\| R_{N} \| b_{6} = θ^{10} / 10! .$

When $θ = 1,$

R_{5} \leq 1 / 10! \approx 2.75 \times 10^{−7} .

When $θ = π / 6,$

R_{5} \leq {(π / 6)}^{10} / 10! \approx 4.26 \times 10^{−10} .

When $θ = π,$

R_{5} \leq π^{10} / 10! = 0.0258 .

The formula $sin θ = θ - \frac{θ^{3}}{3!} + \frac{θ^{5}}{5!} - \frac{θ^{7}}{7!} + ⋯$

will be derived in the next chapter. Use the remainder $\| R_{N} \| \leq b_{N + 1}$

to find a bound for the error in estimating $sin θ$

by the fifth partial sum $θ - θ^{3} / 3! + θ^{5} / 5! − θ^{7} / 7! + θ^{9} / 9!$

for $θ = 1,$

θ = π / 6,

and $θ = π .$

How many terms in $cos θ = 1 - \frac{θ^{2}}{2!} + \frac{θ^{4}}{4!} - \frac{θ^{6}}{6!} + ⋯$

are needed to approximate $cos 1$

accurate to an error of at most $0.00001 ?$

Let $b_{n} = 1 / (2 n - 2)! .$

Then $R_{N} \leq 1 / (2 N)! < 0.00001$

when $(2 N)! > 10^{5}$

or $N = 5$

and $1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \frac{1}{8!} = 0.540325 …,$

whereas $cos 1 = 0.5403023 …$

How many terms in $sin θ = θ - \frac{θ^{3}}{3!} + \frac{θ^{5}}{5!} - \frac{θ^{7}}{7!} + ⋯$

are needed to approximate $sin 1$

accurate to an error of at most $0.00001 ?$

Sometimes the alternating series $\sum_{n = 1}^{\infty} {(−1)}^{n - 1} b_{n}$

converges to a certain fraction of an absolutely convergent series $\sum_{n = 1}^{\infty} b_{n}$

at a faster rate. Given that $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6},$

find $S = 1 - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + ⋯ .$

Which of the series $6 \sum_{n = 1}^{\infty} \frac{1}{n^{2}}$

and $S \sum_{n = 1}^{\infty} \frac{{(−1)}^{n - 1}}{n^{2}}$

gives a better estimation of $π^{2}$

using $1000$

terms?

Let $T = \sum \frac{1}{n^{2}} .$

Then $T - S = \frac{1}{2} T,$

so $S = T / 2 .$

\sqrt{6 \times \sum_{n = 1}^{1000} 1 / n^{2}} = 3.140638 …;

\sqrt{12 \times \sum_{n = 1}^{1000} {(−1)}^{n - 1} / n^{2}} = 3.141591 …;

π = 3.141592 … .

The alternating series is more accurate for $1000$

terms.

The following alternating series converge to given multiples of $π .$

Find the value of $N$

predicted by the remainder estimate such that the $N th$

partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum $N$

for which the error bound holds, and give the desired approximate value in each case. Up to $15$

decimals places, $π = 3.141592653589793 … .$

[T] $\frac{π}{4} = \sum_{n = 0}^{\infty} \frac{{(−1)}^{n}}{2 n + 1},$

error $< 0.0001$

[T] $\frac{π}{\sqrt{12}} = \sum_{k = 0}^{\infty} \frac{{(−3)}^{− k}}{2 k + 1},$

error $< 0.0001$

N = 6,

S_{N} = 0.9068

[T] The series $\sum_{n = 0}^{\infty} \frac{sin (x + π n)}{x + π n}$

plays an important role in signal processing. Show that $\sum_{n = 0}^{\infty} \frac{sin (x + π n)}{x + π n}$

converges whenever $0 < x < π .$

(Hint: Use the formula for the sine of a sum of angles.)

[T] If $\sum_{n = 1}^{N} {(−1)}^{n - 1} \frac{1}{n} \to ln 2,$

what is $1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{8} - \frac{1}{10} - \frac{1}{12} + ⋯ ?$

ln (2) .

The $3 n th$

partial sum is the same as that for the alternating harmonic series.

[T] Plot the series $\sum_{n = 1}^{100} \frac{cos (2 π n x)}{n}$

for $0 \leq x < 1 .$

Explain why $\sum_{n = 1}^{100} \frac{cos (2 π n x)}{n}$

diverges when $x = 0, 1 .$

How does the series behave for other $x ?$

[T] Plot the series $\sum_{n = 1}^{100} \frac{sin (2 π n x)}{n}$

for $0 \leq x < 1$

and comment on its behavior

The series jumps rapidly near the endpoints. For $x$

away from the endpoints, the graph looks like $π (1 / 2 - x) .$

This shows a function in quadrants 1 and 4 that begins at (0, 0), sharply increases to just below 1.5 close to the y axis, decreases linearly, crosses the x axis at 0.5, continues to decrease linearly, and sharply increases just before 1 to 0.

[T] Plot the series $\sum_{n = 1}^{100} \frac{cos (2 π n x)}{n^{2}}$

for $0 \leq x < 1$

and describe its graph.

[T] The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form $\sum_{n = 1}^{\infty} \frac{S_{n}}{n},$

where $s_{n}$

is a randomly generated sequence of $\pm 1's$

in which the values $\pm 1$

are equally likely to occur. Use a random number generator to produce $1000$

random $\pm 1 s$

and plot the partial sums $S_{N} = \sum_{n = 1}^{N} \frac{s_{n}}{n}$

of your random harmonic sequence for $N = 1$

to $1000 .$

Compare to a plot of the first $1000$

partial sums of the harmonic series.

Here is a typical result. The top curve consists of partial sums of the harmonic series. The bottom curve plots partial sums of a random harmonic series.* * *

This shows two curves. The top is an increasing concave down curve. The bottom is a jagged, random harmonic series plot that stays close to 0.

[T] Estimates of $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$

can be accelerated by writing its partial sums as $\sum_{n = 1}^{N} \frac{1}{n^{2}} = \sum_{n = 1}^{N} \frac{1}{n (n + 1)} + \sum_{n = 1}^{N} \frac{1}{n^{2} (n + 1)}$

and recalling that $\sum_{n = 1}^{N} \frac{1}{n (n + 1)} = 1 - \frac{1}{N + 1}$

converges to one as $N \to \infty .$

Compare the estimate of $π^{2} / 6$

using the sums $\sum_{n = 1}^{1000} \frac{1}{n^{2}}$

with the estimate using $1 + \sum_{n = 1}^{1000} \frac{1}{n^{2} (n + 1)} .$

[T] The Euler transform rewrites $S = \sum_{n = 0}^{\infty} {(−1)}^{n} b_{n}$

as $S = \sum_{n = 0}^{\infty} {(−1)}^{n} 2^{− n - 1} \sum_{m = 0}^{n} (\begin{matrix} n \\ m \end{matrix}) b_{n - m} .$

For the alternating harmonic series, it takes the form $ln (2) = \sum_{n = 1}^{\infty} \frac{{(−1)}^{n - 1}}{n} = \sum_{n = 1}^{\infty} \frac{1}{n 2^{n}} .$

Compute partial sums of $\sum_{n = 1}^{\infty} \frac{1}{n 2^{n}}$

until they approximate $ln (2)$

accurate to within $0.0001 .$

How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate $ln (2) .$

By the alternating series test, $\| S_{n} - S \| \leq b_{n + 1},$

so one needs $10^{4}$

terms of the alternating harmonic series to estimate $ln (2)$

to within $0.0001 .$

The first $10$

partial sums of the series $\sum_{n = 1}^{\infty} \frac{1}{n 2^{n}}$

are (up to four decimals) $0.5000, 0.6250, 0.6667, 0.6823, 0.6885, 0.6911, 0.6923, 0.6928, 0.6930, 0.6931$

and the tenth partial sum is within $0.0001$

of $ln (2) = 0.6931 … .$

[T] In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If $a_{n} \geq 0$

is such that $a_{n} \to 0$

as $n \to \infty$

but $\sum_{n = 1}^{\infty} a_{n}$

diverges, then, given any number $A$

there is a sequence $s_{n}$

of $\pm 1's$

such that $\sum_{n = 1}^{\infty} a_{n} s_{n} \to A .$

Show this for $A > 0$

as follows.

Recursively define $s_{n}$
by
$s_{n} = 1$
if
$S_{n - 1} = \sum_{k = 1}^{n - 1} a_{k} s_{k} < A$
and
$s_{n} = −1$
otherwise.
Explain why eventually $S_{n} \geq A,$
and for any
$m$
larger than this
$n,$ $A - a_{m} \leq S_{m} \leq A + a_{m} .$
Explain why this implies that $S_{n} \to A$
as
$n \to \infty .$

Alternating Series

The Alternating Series Test

Proof

Remainder of an Alternating Series

Absolute and Conditional Convergence

Proof

Key Concepts

Key Equations

Glossary