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Alternating Series – Definition, Convergence, and Sum

This article focuses on alternating series – a unique series because of their terms’ unique behavior. Objects that follow oscillatory motion are great examples of values that may have alternating signs or directions. It’ll be interesting to see how we can apply our previous knowledge of series to manipulate and “tame” the different alternating series.

An alternating series is a series where the signs of its terms are alternating between positive and negative signs.

We’ve previously learned about different series where the signs do not vary, such as the arithmetic and harmonic series. In this article, we’ll focus on series that has terms that alternate between positive and negative signs.

  • We’ll understand how we can express these types of series in different forms.

  • We’ll observe alternating series and see if they are convergent.

  • We’ll also learn how to add up the terms of a given alternating series.

Why don’t we begin by understanding the different components of alternating series?

What is an alternating series?                

As we have mentioned in the introduction, alternating sequences or series are unique because the signs of their terms do not remain the same. In fact, they alternate between having negative and positive values.

\begin{aligned}-1, 2, -3, 4, -5, 6, …\end{aligned}

The sequence shown above is a straightforward example of an alternating sequence. We can see that that the odd-numbered terms are negative while even-numbered terms are positive. How do we express this in terms of the nth term of the sequence?

\begin{aligned}-1&= (-1)^1 \cdot 2\\ 2 &= (-1)^2\cdot 2\\ -3 &= (-1)^3 \cdot 3\\ 4 &= (-1)^4 \cdot 4\\.\\.\\.\\a_n &= (-1)^n \cdot n\end{aligned}

From this, we can see that the $n$th term of the sequence can be expressed as $(-1)^n \cdot n$.

The only thing different about alternating sequence and series is that the alternating series represents the sum of an alternating sequence’s terms.

\begin{aligned}-1 + 2 + -3 + 4 + -5 + 6+ … + (-1)^n \cdot n\end{aligned}

Using the example shown for the alternating sequence, it becomes an alternating series once we find the sum of the terms instead.

Alternating series definition

A series, $a_1+ a_2+ a_3+ a_4+ …+ a_n$, is said to be alternating when $a_1$ and $a_2$ have different signs. Similarly, $a_3$ and $a_4$ as well as $a_{n-1}$ and $a_n$ will have alternating signs.

That’s what makes an alternating series – when expressed in terms of $n$, we usually utilize $(-1)^n$ to set the signs of the $n$th term. To understand the reason behind this, let’s observe what happens with $(-1)^n$ when $n$ is even and when $n$ is odd.

Value of $\boldsymbol{n}$

$\boldsymbol{(-1)^n}$

Value of $\boldsymbol{n}$

$\boldsymbol{(-1)^n}$

$1$

$(-1)^1 = -1$

$101$

$(-1)^{101} = -1$

$2$

$(-1)^2= 1$

$102$

$(-1)^{102}= 1$

$3$

$(-1)^3= -1$

$103$

$(-1)^{103}= -1$

$4$

$(-1)^4= 1$

$104$

$(-1)^{104}= 1$

As we can see, when $n$ is even, $(-1)^n$ is equal to $1$. Similarly, when $n$ is odd, we expect $(-1)^n$ to be equal to $-1$.

Here are the two possible forms of an alternating series:

\begin{aligned}\sum_{n=1}^{\infty} (-1)^n a_n &= -a_1 + a_2 – a_3 + a_4 -…\\\\\sum_{n=1}^{\infty} (-1)^{n + 1} a_n &= a_1 – a_2 + a_3 – a_4 + … \end{aligned}

Keep in mind that these general forms apply when $a_n \geq 0$ and $n$ is a positive integer.

How to find the sum of an alternating series?                

When asked to add alternating series, there are different approaches we can apply.

  • The most convenient approach identifies whether the alternating series is a type of arithmetic, harmonic, or geometric series.

  • When they are, we can then apply the properties we’ve learned about the series so that we can immediately find the sum of the given alternating series.

  • We can also separate the negative terms and the positive terms and then combine their respective sums.

  • In case the series is challenging to manipulate, we can also estimate the sum of an alternating series by extending the alternating series test.

Rewriting the Alternating Series

Let’s say we have $-2 + 4 – 6 + 8 – 10 + …. -50$. We can find its sum by rewriting the series as two sets of regular arithmetic series.

First Arithmetic Series

Second Arithmetic Series

$-2 – 6 – 10 – … -50$

$4+ 8 +12 +… +48$

The first series has a common difference of $-4$ and a leading term, $-2$. Since there are $12$ terms, the sum of the first series, $S_1$, is shown below.

\begin{aligned} S_1 &= \dfrac{13}{2} (-2-50)\\&= 13(-52)\\&= -338 \end{aligned}

We’ll apply a similar process to find the sum of the second arithmetic series.

\begin{aligned} S_2 &= \dfrac{12}{2} (4 + 48)\\&= 6(52)\\&= 312 \end{aligned}

Adding the two sums will give us the sum of the original series.

\begin{aligned} S_1 + S_2 &= -338 + 312\\&= -26 \end{aligned}

Another way of rewriting the alternating series is to group each pair of consecutive terms, which will leave $50$ unpaired. But let’s see what happens, shall we?

\begin{aligned} -2 + 4 – 6 + 8 – 10 + …. -50 &= (-2 + 4) + (– 6 + 8) + (– 10 + 12) +….+(-46 + 48) -50 \\&= 2 + 2 + 2 +…+ 2 – 50\\&=12(2) – 50\\&= 24 – 50\\&= -26 \end{aligned}

We combined each pair and left $-50$ unpaired. Since each of the twelve pairs shares a common sum of $2$, we can then find the sum of all $24$ terms before $-50$.

The two methods show that whenever possible, we can manipulate alternating series actually to find their sum.

The problem, however, arises when it’s not easy for us to manipulate the series. When this happens, we can resort to estimating the sum of the alternating series instead. This method is most helpful when we’re working with infinite alternating series.

Applying the Alternating Series Test

A quick recap on the alternating series test: when $a_n$ is a positive term that is continuously decreasing as the series progresses or simply, $\lim_{n \rightarrow \infty} a_n = 0$, the alternating series converges.

This condition applies to both general forms of an alternating series: $\sum_{n=1}^{\infty} (-1)^n a_n$ and $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$.

How does this apply to our estimating the sum of an alternating series?

Let’s say our alternating series satisfies and meets the alternating series test conditions; a good estimate of the alternating series’ sum, $S$, is by finding the partial sum of the first $n$ terms.

Let’s say if we want to estimate the sum of $\sum_{n = 1}^{\infty} (-1)^{n + 1}\dfrac{1}{n^4}$ to two decimal places, we find the first consecutive partial sums where the first two decimal places have identical digits.

Terms of the Alternating Series

Partial Sum, $\boldsymbol{S_n}$

$1$

$S_1=1$

$1 – \dfrac{1}{2^4}$

$S_2 = 0.9375$

$1 – \dfrac{1}{2^4} + \dfrac{1}{3^4}$

$S_3 \approx 0.9498$

$1 – \dfrac{1}{2^4} + \dfrac{1}{3^4} – \dfrac{1}{4^4}$

$S_4 \approx 0.9459$

$1 – \dfrac{1}{2^4} + \dfrac{1}{3^4} – \dfrac{1}{4^4}+ \dfrac{1}{5^4}$

$S_5 \approx 0.9475$

From this, we can see that a good estimate (in two decimal places) of the alternate series is $0.94$.

But if we want a more accurate answer or are required to estimate the sum at a specific error margin, it’s best that we find the remainder of the series.

The remainder (or truncation error), $R_n$, of an alternating series indicates the amount of error we’re allowed in the approximation. Let’s say we have $S_n$ represent the alternating series’ partial sum and $S$ represents the sum of the entire series, we have the following relationship:

\begin{aligned}\left| R_n\right| &= |S – S_n|\\&\leq |S_{n+1} – S_n| \\&= a_{n +1} \end{aligned}

This means that we estimate the sum of the alternating series by finding the value of $S_n$ where it is less than or equal to the next term, $a_{n+1}$.

Let’s say we want to estimate the value of $\sum_{n = 1}^{\infty} (-1)^{n} \dfrac{1}{n}$ so that the error is less than $0.1$.

We want to find the truncation error or remainder, $\left| R_n\right|$, to be less that $0.01$ or $10^{-1}$. This means that we need to find the value of $n$ where $\left| R_n\right| = a_{n+1} < 10^{-1}$.

\begin{aligned} a_{n+1}&\leq 10^{-1} \\\dfrac{1}{n + 1}&\leq 10^{-1}\\\dfrac{1}{n+1} &\leq \dfrac{1}{10}\\n + 1&\geq 10 \\n &\geq9 \end{aligned}

This means we can use the partial sum of the first $n = 9$ terms to estimate the alternating series’s sum and be guaranteed that the error is not greater than $0.1$.

\begin{aligned} \sum_{n = 1}^{9} (-1)^{n} \dfrac{1}{n} &= -1 + \dfrac{1}{2} -\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+ \dfrac{1}{6} -\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}\\&= -\dfrac{1879}{2520} \\&\approx -0.7456\end{aligned}

This sum is less than $0.1$ as well – showing that this is a good estimate of the series’ sum given the approximation error that we’re allowed.

We’ve discussed extensively alternating series, so let’s apply everything we’ve learned so far by answering the sample problems shown below.

Example 1

Write the following alternating series in terms of the summation notation in which we begin with $n = 1$.

a. $3 – 6 + 9 – 12 + 15 – …$

b. $1 – \dfrac{1}{4} + \dfrac{1}{9} – \dfrac{1}{16} +…$

 c. $-1 + \dfrac{2}{3} – \dfrac{2}{4} + \dfrac{2}{5} – …$

d. $- \dfrac{4}{6} + \dfrac{5}{7} – \dfrac{6}{8} + \dfrac{7}{9}-…$

Solution

Let’s observe how each term progresses so that we can find a pattern for the alternating series. Why don’t we begin with the first series?

\begin{aligned}3 &= (-1)^2 \cdot (3 \cdot 1)\\-6 &= (-1)^3 \cdot (3 \cdot 2)\\9 &= (-1)^4 \cdot (3 \cdot 3)\\-12 &= (-1)^5 \cdot (3 \cdot 4)\\15 &= (-1)^6 \cdot (3 \cdot 5)\\.\\.\\. a_n &= (-1)^{n + 1} \cdot (3 \cdot n)\end{aligned}

By observing these terms, we can see that the $n$th term can be expressed as $(-1)^{n + 1} 3n$.

a. This means that the the alternating series can expressed as $\sum_{n = 1}^{\infty} (-1)^{n + 1} 3n$.

We’ll apply a similar process to find the expression for the $n$th term.

\begin{aligned} 1 &= (-1)^{2} \dfrac{1}{1^2}\\ -\dfrac{1}{4} &= (-1)^{3} \dfrac{1}{2^2}\\\dfrac{1}{9} &= (-1)^{4} \dfrac{1}{3^2}\\-\dfrac{1}{16} &= (-1)^{5} \dfrac{1}{4^2}\\.\\.\\.\\a_n &= (-1)^{n + 1}\dfrac{1}{n^2}\end{aligned}

b. Hence, the second alternating series can be expressed as $\sum_{n = 1}^{\infty} (-1)^{n + 1} \dfrac{1}{n^2}$.

We can now take a look at the third alternating series. This series looks familiar, right? That’s because this is an alternating harmonic series.

\begin{aligned} -1 &= (-1)^{1} \left(2 \cdot \dfrac{1}{1 + 1}\right)\\ \dfrac{2}{3} &= (-1)^{2} \left(2 \cdot \dfrac{1}{2 + 1}\right)\\-\dfrac{2}{4} &= (-1)^{3} \left(2 \cdot \dfrac{1}{3 + 1}\right)\\\dfrac{2}{5} &= (-1)^{4} \left(2 \cdot \dfrac{1}{4 + 1}\right)\\.\\.\\.\\a_n &= (-1)^{n} \left(2 \cdot \dfrac{1}{n + 1}\right)\end{aligned}

c. This means that the summation form of the series is $\sum_{n = 1}^{\infty} (-1)^{n } \dfrac{2}{n + 1}$.

Let’s work on the fourth alternating series now – take note of the solution since finding the $n$th term can get tricky.

\begin{aligned} -\dfrac{4}{6}&= (-1)^{1} \cdot \left(\dfrac{1 + 3}{1 + 5}\right)\\ \dfrac{5}{7} &= (-1)^{2} \cdot \left(\dfrac{2 + 3}{2 + 5}\right)\\-\dfrac{6}{8} &= (-1)^{3} \cdot \left(\dfrac{3 + 3}{3 + 5}\right)\\\dfrac{7}{9} &= (-1)^{4} \cdot \left(\dfrac{4 + 3}{4 + 5}\right)\\.\\.\\.\\a_n &= (-1)^{n} \left( \dfrac{n + 3}{n + 5}\right)\end{aligned}

d. Hence, the summation form of the alternating series is $\sum_{n = 1}^{\infty} (-1)^{n } \dfrac{n + 3}{n + 5}$.

Example 2

Find the sum of the following alternating series.

    a. $-3 + 6 – 9 + 12 – … + 36$

    b. $-1 + \dfrac{1}{2} – \dfrac{1}{4} + \dfrac{1}{8} – … -\dfrac{1}{1024}$

Solution

The first series is an alternating arithmetic series, so we can group the terms as pairs and check out each pair’s sum. We can then multiply the individual sum by the number of pairs.

\begin{aligned}-3 + 6 – 9 + 12 – … + 36 &= (-3 + 6) + (– 9 + 12) + …+(-27 + 36)\\&= 3 + 3 + … + 3\\&=6(3)\\ &= 18\end{aligned}

a. This means that $-3 + 6 – 9 + 12 – … + 36 = 18$.

Why don’t we write the denominators of the next series as powers of two?

\begin{aligned}-1 + \dfrac{1}{2} – \dfrac{1}{4} + \dfrac{1}{8} – … -\dfrac{1}{1024} &= -1 + \dfrac{1}{2^1} – \dfrac{1}{2^2} + \dfrac{1}{2^3} – … -\dfrac{1}{2^{10}}\end{aligned}

In fact, this series is an alternating geometric – the alternating sign is due to the sign of the common ratio, $-\dfrac{1}{2}$. We can use the geometric series formula, $S_n = \dfrac{a(r^n – 1)}{r – 1}$, where $r = -\dfrac{1}{2}$ and $n = 11$.

\begin{aligned}S_{11} &= \dfrac{-1\left[-\left(\dfrac{1}{2}\right)^{11}- 1\right]}{-\dfrac{1}{2} – 1}\\&= \dfrac{-1\left(-\dfrac{1}{2048} – 1 \right )}{-\dfrac{3}{2}}\\&= \dfrac{\dfrac{2049}{2048}}{-\dfrac{3}{2}}\\&= \dfrac{2049}{2048} \cdot – \dfrac{2}{3}\\&= – \dfrac{683}{2048}\end{aligned}

b. This means that the sum of the first eleven terms of the series is $- \dfrac{683}{2048}$.

Example 3

Let’s say we have the alternating series, $\sum_{n = 1}^{\infty} (-1)^{n + 1 } \dfrac{1}{n ^3}$.

a. Estimate the sum of the series to the two nearest decimal places.

b. What is a good approximation for the sum if we want a truncation error of at most $0.001$?

Solution

To estimate the sum of the alternating series, let’s inspect the first $n$ partial sums’ values until we have two consecutive ones with identical digits on their decimal places.

Terms of the Alternating Series

Partial Sum, $\boldsymbol{S_n}$

$1$

$S_1=1$

$1 – \dfrac{1}{2^3}$

$S_2 = 0.875$

$1 – \dfrac{1}{2^3} + \dfrac{1}{3^3}$

$S_3 \approx 0.912$

$1 – \dfrac{1}{2^3} + \dfrac{1}{3^3} – \dfrac{1}{4^3}$

$S_4 \approx 0.896$

$1 – \dfrac{1}{2^3} + \dfrac{1}{3^3} – \dfrac{1}{4^3}+ \dfrac{1}{5^3}$

$S_5 \approx 0.904$

$1 – \dfrac{1}{2^3} + \dfrac{1}{3^3} – \dfrac{1}{4^3}+ \dfrac{1}{5^3} – \dfrac{1}{6^3}$

$S_6 \approx 0.900$

We can see that the first five terms’ partial sums and the first six terms are approximately $0.90$.

a. This means that the estimated sum of the alternating series is $0.90$.

For the second condition, we’ll need to find the value of $n$ where $\left| R_n\right| = a_{n+1} < 10^{-3}$.

\begin{aligned} a_{n+1}&\leq 10^{-3} \\\dfrac{1}{(n + 3)^3}&\leq 10^{-3}\\\dfrac{1}{(n+1)^3} &\leq \dfrac{1}{10}\\(n + 1)^3&\geq 10^3 \\n + 1 &\geq 10 \\ n&\geq 9 \end{aligned}

Use the partial sum of the first $n = 9$ terms to estimate the alternating series’s sum and be guaranteed that the error is not greater than $0.001$.

\begin{aligned} \sum_{n = 1}^{9} (-1)^{n} \dfrac{1}{n^3} &= -1 + \dfrac{1}{2^3} -\dfrac{1}{3^3}+\dfrac{1}{4^3}-\dfrac{1}{5^3}+ \dfrac{1}{6^3} -\dfrac{1}{7^3}+\dfrac{1}{8^3}-\dfrac{1}{9^3}\\&\approx -0.9021\end{aligned}

We can confirm that the sum is less than $0.001$.

b. Hence, a good estimate of the sum is $-0.9021$.

Practice Questions

1. Write the following alternating series in terms of the summation notation in which we begin with $n = 1$.

    a. $4 – 8 + 12 – 16+ 20 – …$

    b. $3 – \dfrac{3}{4} + \dfrac{1}{3} – \dfrac{3}{16} + …$

    c. $-1 + \dfrac{1}{3} – \dfrac{1}{5} + \dfrac{1}{7} – …$

    d. $– 3 + 12 – 27 + 48…$

2. Find the sum of the following alternating series.

    a. $-4+ 8 – 12 + 16 – … + 80$

    b. $-1 + \dfrac{1}{3} – \dfrac{1}{9} + \dfrac{1}{27} – … -\dfrac{1}{3^10}$

3.Let’s say we have the alternating series, $\sum_{n = 1}^{\infty} (-1)^{n + 1 } \dfrac{1}{n ^5}$.

a. Estimate the sum of the series to the two nearest decimal places.

b. What is a good approximation for the sum if we want a truncation error of at most $0.00001$?

Answer Key

1.

a. $\sum_{n=1}^{\infty} (-1)^{n+1} 4n$

b. $\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{3}{n^2}$

c. $\sum_{n=1}^{\infty} (-1)^n\dfrac{1}{2n -1}$

d. $\sum_{n=1}^{\infty} (-1)^n 3n^2$

2.

a. $40$

b. $-\dfrac{44287}{59049}$

3.

a. $0.97$

b. $0.972126$

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