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## Section7.5Ratio Test and Alternating Series

###### Motivating Questions
• How can we test convergence without a simple integral or known series to compare to?

• What is an alternating series?

• Under what conditions does an alternating series converge? Why?

So far, we've mostly considered series with exclusively nonnegative terms that have straightforward comparisons. In this section, we will consider one powerful test to compare many series to geometric series using the ratios of consecutive terms. Also, we consider series that have some negative terms.

### SubsectionThe Ratio Test

The Limit Comparison Test works well if we can find a series with known behavior to compare. But such series are not always easy to find. Below, we will examine a test that allows us to consider the behavior of a series by comparing it to a geometric series, without knowing in advance which geometric series we need.

###### Example7.32

Consider the series defined by

\begin{equation} \sum_{k=1}^{\infty} \frac{2^k}{3^k-k}\text{.}\label{eq-8-3-ratio-example}\tag{7.14} \end{equation}

This series is not a geometric series, but this example will illustrate how we might compare this series to a geometric one. Recall that a series $\sum a_k$ is geometric if the ratio $\frac{a_{k+1}}{a_k}$ is always the same. For the series in (7.14), note that $a_k = \frac{2^k}{3^k-k}\text{.}$

1. To see if $\sum \frac{2^k}{3^k-k}$ is comparable to a geometric series, we analyze the ratios of successive terms in the series. Complete Table7.33, listing your calculations to at least 8 decimal places.

2. Based on your calculations in Table7.33, what can we say about the ratio $\frac{a_{k+1}}{a_k}$ if $k$ is large?

3. Do you agree or disagree with the statement: the series $\sum \frac{2^k}{3^k-k}$ is approximately geometric when $k$ is large? If not, why not? If so, do you think the series $\sum \frac{2^k}{3^k-k}$ converges or diverges? Explain.

1.  $k$ $5$ $10$ $20$ $21$ $22$ $23$ $24$ $25$ $\dfrac{a_{k+1}}{a_k}$ 0.6583679115 0.6665951585 0.6666666642 0.6666666658 0.6666666664 0.6666666666 0.6666666666 0.6666666667
2. $\frac{a_{k+1}}{a_k} \approx \frac{2}{3}$ when $k$ is large.

3. $\sum \frac{2^k}{3^k-k}$ converges.

Solution
1. Ratios of successive summands in the series are shown (to 10 decimal places) in the following table.

 $k$ $5$ $10$ $20$ $21$ $22$ $23$ $24$ $25$ $\dfrac{a_{k+1}}{a_k}$ 0.6583679115 0.6665951585 0.6666666642 0.6666666658 0.6666666664 0.6666666666 0.6666666666 0.6666666667
2. The calculations in the table in part (a) seem to indicate that the ratio $\frac{a_{k+1}}{a_k}$ is roughly $\frac{2}{3}$ when $k$ is large.

3. Since $\frac{a_{k+1}}{a_k} \approx \frac{2}{3}$ for large $k\text{,}$ the series $\sum \frac{2^k}{3^k-k}$ is approximately the same as $\sum \left(\frac{2}{3}\right)^k$ when $k$ is large. So the series $\sum \frac{2^k}{3^k-k}$ is approximately geometric with ratio $\frac{2}{3}$ when $k$ is large. Since the series $\sum \left(\frac{2}{3}\right)^k$ converges because the ratio is less than 1, we expect that the series $\sum \frac{2^k}{3^k-k}$ will also converge.

We can generalize the argument in Example7.32 in the following way. Consider the series $\sum a_k\text{.}$ If

\begin{equation*} \frac{a_{k+1}}{a_k} \approx r \end{equation*}

for large values of $k\text{,}$ then $a_{k+1} \approx ra_k$ for large $k$ and the series $\sum a_k$ is approximately the geometric series $\sum ar^k$ for large $k\text{.}$ Since the geometric series with ratio $r$ converges only for $-1 \lt r \lt 1\text{,}$ we see that the series $\sum a_k$ will converge if

\begin{equation*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = r \end{equation*}

for a value of $r$ such that $|r| \lt 1\text{.}$ This result is known as the Ratio Test.

###### The Ratio Test

Let $\sum a_k$ be an infinite series. Suppose

\begin{equation*} \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_k|} = r\text{.} \end{equation*}
1. If $0 \leq r \lt 1\text{,}$ then the series $\sum a_k$ converges.

2. If $1 \lt r\text{,}$ then the series $\sum a_k$ diverges.

3. If $r = 1\text{,}$ then the test is inconclusive.

Note well: The Ratio Test looks at the limit of the ratio of consecutive terms of a given series; in so doing, the test is asking, is this series approximately geometric? If so, the test uses the limit of the ratio of consecutive terms to determine if the given series converges.

We have now encountered several tests for determining convergence or divergence of series.

• The Divergence Test can be used to show that a series diverges, but never to prove that a series converges.
• We used the Integral Test to determine the convergence status of an entire class of series, the $p$-series. More generally, the Integral Test may be applied whenever the series terms correspond to a function that is positive and decreasing and whose convergence behavior when integrated is known.
• The Direct Comparison Test and Limit Comparison Test work well for series that involve rational functions and which can therefore be compared to $p$-series and other series whose convergence behavior we know.
• Finally, the Ratio Test allows us to compare our series to a geometric series; it is particularly useful for series that involve $n$th powers and factorials.
• Another test, called the Root Test, is discussed in the exercises.

One of the challenges of determining whether a series converges or diverges is finding which test answers that question.

###### Example7.34

Determine whether each of the following series converges or diverges. Explicitly state which test you use.

1. $\sum \frac{k}{2^k}$

2. $\sum \frac{k^3+2}{k^2+1}$

3. $\sum \frac{10^k}{k!}$

4. $\sum \frac{k^3-2k^2+1}{k^6+4}$

1. $\sum \frac{k}{2^k}$ converges.

2. $\sum \frac{k^3+2}{k^2+1}$ diverges.

3. $\sum \frac{10^k}{k!}$ converges.

4. $\sum \frac{k^3-2k^2+1}{k^6+4}$ converges.

Solution
1. Since $\lim_{n \to \infty} \frac{n}{2^n} = 0\text{,}$ the Divergence Test does not apply. The Ratio Test is a good one to apply with series whose terms involve exponentials and polynomials.

In this example we have $a_{n+1} = \frac{n+1}{2^{n+1}}$ and $a_n = \frac{n}{2^n}\text{.}$ So

\begin{equation*} \frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \frac{n+1}{2n}\text{.} \end{equation*}

As $n$ goes to infinity, the fraction $\frac{n+1}{n}$ goes to 1 and so

\begin{equation*} \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2}\text{.} \end{equation*}

So in this example, for large $n$ we have that $\frac{a_{n+1}}{a_n} \approx \frac{1}{2}$ or that $\sum a_n$ is roughly a geometric series for large $n$ with ratio $\frac{1}{2}\text{.}$ Since geometric series with ratio $\frac{1}{2}$ converges, we can conclude that

\begin{equation*} \sum \frac{n}{2^n} \end{equation*}

converges as well.

2. For this series, notice that the degree of the numerator is greater than the degree of the denominator, so

\begin{equation*} \lim_{n \to \infty} \frac{n^3+2}{n^2+1} = \infty\text{.} \end{equation*}

Since this limit is not 0, the Divergence Test shows that $\sum \frac{k^3+2}{k^2+1}$ diverges.

3. We use the Ratio Test here with $a_k = \frac{10^k}{k!}\text{.}$ Now

\begin{align*} \lim_{k \to \infty} \frac{a_{k+1}}{a_k} \amp = \lim_{k \to \infty} \frac{ \frac{10^{k+1}}{(k+1)!} }{ \frac{10^k}{k!} }\\ \amp = \lim_{k \to \infty} \frac{10^{k+1}(k!)}{10^k(k+1)!}\\ \amp = \lim_{k \to \infty} \frac{10}{k+1}\\ \amp = 0\text{,} \end{align*}

so the Ratio Test shows that $\sum \frac{10^k}{k!}$ converges.

4. If we ignore the highest powered terms, then the series $\sum \frac{k^3-2k^2+1}{k^6+4}$ looks like the series $\sum \frac{k^3}{k^6} = \sum \frac{1}{k^3}\text{.}$ We formally apply the Limit Comparison Test as follows:

\begin{align*} \lim_{k \to \infty} \frac{ \frac{k^3-2k^2+1}{k^6+4} }{ \frac{1}{k^3} } \amp = \lim_{k \to \infty} \frac{k^6-2k^5+k^3}{k^6+4}\\ \amp = \lim_{k \to \infty} \frac{1 - \frac{2}{k} + \frac{1}{k^3}}{1 + \frac{4}{k^6}}\\ \amp 1\text{.} \end{align*}

Since $\sum \frac{1}{k^3}$ is a $p$-series with $p=3 \gt 1\text{,}$ the series $\sum \frac{1}{k^3}$ converges. Consequently, the Limit Comparison Test shows that $\sum \frac{k^3-2k^2+1}{k^6+4}$ converges as well.

### SubsectionThe Alternating Series Test

Consider the following geometric series:

\begin{equation*} 2 - \frac{4}{3} + \frac{8}{9} - \cdots + 2 \left(-\frac{2}{3} \right)^n + \cdots\text{.} \end{equation*}

Notice that $a = 2$ and $r = -\frac{2}{3}\text{,}$ so that every other term alternates in sign. This series converges to

\begin{equation*} S = \frac{a}{1-r} = \frac{2}{1- \left(-\frac{2}{3}\right)} = \frac{6}{5}\text{.} \end{equation*}

In Example7.35 and our following discussion, we investigate the behavior of similar series where consecutive terms have opposite signs.

###### Example7.35

Example7.20 showed how we can approximate the number $e$ with linear, quadratic, and other polynomial approximations. We use a similar approach in this example to obtain linear and quadratic approximations to $\ln(2)\text{.}$ Along the way, we encounter a type of series that is different than most of the ones we have seen so far. Throughout this example, let $f(x) = \ln(1+x)\text{.}$

1. Find the tangent line to $f$ at $x=0$ and use this linearization to approximate $\ln(2)\text{.}$ That is, find $L(x)\text{,}$ the tangent line approximation to $f(x)\text{,}$ and use the fact that $L(1) \approx f(1)$ to estimate $\ln(2)\text{.}$

2. The linearization of $\ln(1+x)$ does not provide a very good approximation to $\ln(2)$ since $1$ is not that close to $0\text{.}$ To obtain a better approximation, we alter our approach; instead of using a straight line to approximate $\ln(2)\text{,}$ we use a quadratic function to account for the concavity of $\ln(1+x)$ for $x$ close to $0\text{.}$ With the linearization, both the function's value and slope agree with the linearization's value and slope at $x=0\text{.}$ We will now make a quadratic approximation $P_2(x)$ to $f(x) = \ln(1+x)$ centered at $x=0$ with the property that $P_2(0) = f(0)\text{,}$ $P'_2(0) = f'(0)\text{,}$ and $P''_2(0) = f''(0)\text{.}$

1. Let $P_2(x) = x - \frac{x^2}{2}\text{.}$ Show that $P_2(0) = f(0)\text{,}$ $P'_2(0) = f'(0)\text{,}$ and $P''_2(0) = f''(0)\text{.}$ Use $P_2(x)$ to approximate $\ln(2)$ by using the fact that $P_2(1) \approx f(1)\text{.}$

2. We can continue approximating $\ln(2)$ with polynomials of larger degree whose derivatives agree with those of $f$ at $0\text{.}$ This makes the polynomials fit the graph of $f$ better for more values of $x$ around $0\text{.}$ For example, let $P_3(x) = x - \frac{x^2}{2}+\frac{x^3}{3}\text{.}$ Show that $P_3(0) = f(0)\text{,}$ $P'_3(0) = f'(0)\text{,}$ $P''_3(0) = f''(0)\text{,}$ and $P'''_3(0) = f'''(0)\text{.}$ Taking a similar approach to preceding questions, use $P_3(x)$ to approximate $\ln(2)\text{.}$

3. If we used a degree $4$ or degree $5$ polynomial to approximate $\ln(1+x)\text{,}$ what approximations of $\ln(2)$ do you think would result? Use the preceding questions to conjecture a pattern that holds, and state the degree $4$ and degree $5$ approximation.

Solution
1. The linearization of $f$ at $x=a$ is

\begin{equation*} f(a) + f'(a)(x-a)\text{,} \end{equation*}

so the linearization $P_1(x)$ of $f(x) = \ln(1+x)$ at $x=0$ is

\begin{equation*} P_1(x) = 0 + \frac{1}{1+0}(x-0) = x\text{.} \end{equation*}

Now

\begin{equation*} f(x) \approx P_1(x) \end{equation*}

for $x$ close to $0$ and so

\begin{equation*} \ln(2) = \ln(1+1) \approx P_1(1) = 1\text{.} \end{equation*}
1. The derivatives of $P_2$ and $f$ are

\begin{align*} P_2(x) \amp = x - \frac{x^2}{2} \amp f(x) \amp = \ln(1+x)\\ P'_2(x) \amp = 1-x \amp f'(x) \amp = \frac{1}{1+x}\\ P''_2(x) \amp = -1 \amp f''(x) \amp = -\frac{1}{(1+x)^2}\text{,} \end{align*}

and so the derivatives of $P_2$ and $f$ evaluated at 0 are

\begin{align*} P_2(0) \amp = 0 \amp f(0) \amp = \ln(1) = 0\\ P'_2(0) \amp = 1 \amp f'(0) \amp = \frac{1}{1+0} = 1\\ P''_2(0) \amp = -1 \amp f''(0) \amp= -\frac{1}{(1+0)^2} = -1\text{.} \end{align*}

Then

\begin{equation*} \ln(2) = \ln(1+1) \approx P_2(1) = 1 - \frac{1}{2} = \frac{1}{2}\text{.} \end{equation*}
2. The derivatives of $P_3$ and $f$ are

\begin{align*} P_3(x) \amp = x-\frac{x^2}{2}+\frac{x^3}{3} \amp f(x) \amp = \ln(1+x)\\ P'_3(x) \amp = 1 - x + x^2 \amp f'(x) \amp = \frac{1}{1+x}\\ P''_3(x) \amp = -1+2x \amp f''(x) \amp = -\frac{1}{(1+x)^2}\\ P'''_3(x) \amp = 2 \amp f'''(x) \amp = \frac{2}{(1+x)^3}\text{,} \end{align*}

and so the derivatives of $P_3$ and $f$ evaluated at 0 are

\begin{align*} P_3(0) \amp = 0 \amp f(0) \amp = \ln(1+0) = 0\\ P'_3(0) \amp = 1 \amp f'(0) \amp = \frac{1}{1+0} = 1\\ P''_3(0) \amp = -1 \amp f''(0) \amp = -\frac{1}{(1+0)^2} = -1\\ P'''_3(0) \amp = 2 \amp f'''(0) \amp = \frac{2}{(1+0)^3} = 2\text{.} \end{align*}

Then

\begin{equation*} \ln(2) = \ln(1+1) \approx P_3(0) = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \approx 0.83\text{.} \end{equation*}

Example7.35 gives us several approximations to $\ln(2)\text{.}$ The linear approximation is $1\text{,}$ and the quadratic approximation is $1 - \frac{1}{2} = \frac{1}{2}\text{.}$ If we continue this process, cubic, quartic (degree $4$), quintic (degree $5$), and higher degree polynomials give us the approximations to $\ln(2)$ in Table7.36.

The pattern here shows that $\ln(2)$ can be approximated by the partial sums of the infinite series

\begin{equation} \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}\label{eq-ln-2}\tag{7.15} \end{equation}

where the alternating signs are indicated by the factor $(-1)^{k+1}\text{.}$ We call such a series an alternating series.

Using computational technology, we find that the sum of the first 100 terms in this series is 0.6881721793. As a comparison, $\ln(2) \approx 0.6931471806\text{.}$ This shows that even though the series(7.15) converges to $\ln(2)\text{,}$ it must do so quite slowly, since the sum of the first 100 terms isn't particularly close to $\ln(2)\text{.}$ We will investigate the issue of how quickly an alternating series converges later in this section.

###### Alternating Series

An alternating series is a series of the form

\begin{equation*} \sum_{k=0}^{\infty} (-1)^k a_k\text{,} \end{equation*}

where $a_k \gt 0$ for each $k\text{.}$

We have some flexibility in how we write an alternating series; for example, the series

\begin{equation*} \sum_{k=1}^{\infty} (-1)^{k+1} a_k\text{,} \end{equation*}

whose index starts at $k = 1\text{,}$ is also alternating. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior.

It is important to remember that most of the series tests we have seen in previous sections apply only to series with nonnegative terms. Alternating series require a different test.

###### Example7.37

Remember that, by definition, a series converges if and only if its corresponding sequence of partial sums converges.

1. Calculate the first few partial sums (to 10 decimal places) of the alternating series

\begin{equation*} \sum_{k=1}^{\infty} (-1)^{k+1}\frac{1}{k}\text{.} \end{equation*}

Label each partial sum with the notation $S_n = \sum_{k=1}^{n} (-1)^{k+1}\frac{1}{k}$ for an appropriate choice of $n\text{.}$

2. Plot the sequence of partial sums from part (a). What do you notice about this sequence?

1.  $\sum_{k=1}^{1} (-1)^{k+1}\frac{1}{k}$ $1$ $\sum_{k=1}^{6} (-1)^{k+1}\frac{1}{k}$ $0.6166666667$ $\sum_{k=1}^{2} (-1)^{k+1}\frac{1}{k}$ $0.5$ $\sum_{k=1}^{7} (-1)^{k+1}\frac{1}{k}$ $0.7595238095$ $\sum_{k=1}^{3} (-1)^{k+1}\frac{1}{k}$ $0.8333333333$ $\sum_{k=1}^{8} (-1)^{k+1}\frac{1}{k}$ $0.6345238095$ $\sum_{k=1}^{4} (-1)^{k+1}\frac{1}{k}$ $0.5833333333$ $\sum_{k=1}^{9} (-1)^{k+1}\frac{1}{k}$ $0.7456349206$ $\sum_{k=1}^{5} (-1)^{k+1}\frac{1}{k}$ $0.7833333333$ $\sum_{k=1}^{10} (-1)^{k+1}\frac{1}{k}$ $0.6456349206$
2. There appears to be a limit for the sequence of partial sums.

Solution
1. The completed table is shown below.

 $\sum_{k=1}^{1} (-1)^{k+1}\frac{1}{k}$ $1$ $\sum_{k=1}^{6} (-1)^{k+1}\frac{1}{k}$ $0.6166666667$ $\sum_{k=1}^{2} (-1)^{k+1}\frac{1}{k}$ $0.5$ $\sum_{k=1}^{7} (-1)^{k+1}\frac{1}{k}$ $0.7595238095$ $\sum_{k=1}^{3} (-1)^{k+1}\frac{1}{k}$ $0.8333333333$ $\sum_{k=1}^{8} (-1)^{k+1}\frac{1}{k}$ $0.6345238095$ $\sum_{k=1}^{4} (-1)^{k+1}\frac{1}{k}$ $0.5833333333$ $\sum_{k=1}^{9} (-1)^{k+1}\frac{1}{k}$ $0.7456349206$ $\sum_{k=1}^{5} (-1)^{k+1}\frac{1}{k}$ $0.7833333333$ $\sum_{k=1}^{10} (-1)^{k+1}\frac{1}{k}$ $0.6456349206$
2. The entries in the sequence of partial sums from part (a) is shown in the figure below.

Notice that the partial sums seem to oscillate back and forth around some fixed number, getting closer to that fixed number at each successive step. So there appears to be a limit for this sequence of partial sums.

Example7.37 illustrates the general behavior of any convergent alternating series. We see that the partial sums of the alternating harmonic series oscillate around a fixed number that turns out to be the sum of the series.

Recall that if $\lim_{k \to \infty} a_k \neq 0\text{,}$ then the series $\sum a_k$ diverges by the Divergence Test. From this point forward, we will thus only consider alternating series

\begin{equation*} \sum_{k=1}^{\infty} (-1)^{k+1} a_k \end{equation*}

in which the sequence $a_k$ consists of positive numbers that decrease to $0\text{.}$ The $n$th partial sum $S_n$ is

\begin{equation*} S_n = \sum_{k=1}^n (-1)^{k+1} a_k\text{.} \end{equation*}

Notice that

• $S_2 = a_1 - a_2\text{,}$ and since $a_1 > a_2$ we have $0 \lt S_2 \lt S_1 \text{.}$

• $S_3 = S_2+a_3$ and so $S_2 \lt S_3\text{.}$ But $a_3 \lt a_2\text{,}$ so $S_3 \lt S_1\text{.}$ Thus, $0 \lt S_2 \lt S_3 \lt S_1 \text{.}$

• $S_4 = S_3-a_4$ and so $S_4 \lt S_3\text{.}$ But $a_4 \lt a_3\text{,}$ so $S_2 \lt S_4\text{.}$ Thus, $0 \lt S_2 \lt S_4 \lt S_3 \lt S_1 \text{.}$

• $S_5 = S_4+a_5$ and so $S_4 \lt S_5\text{.}$ But $a_5 \lt a_4\text{,}$ so $S_5 \lt S_3\text{.}$ Thus, $0 \lt S_2 \lt S_4 \lt S_5 \lt S_3 \lt S_1 \text{.}$

This pattern continues as illustrated in Figure7.39 (with $n$ odd) so that each partial sum lies between the previous two partial sums. Note further that the absolute value of the difference between the $(n-1)$st partial sum $S_{n-1}$ and the $n$th partial sum $S_n$ is

\begin{equation*} \left\lvert S_n - S_{n-1} \right\rvert = a_n\text{.} \end{equation*}

Because the sequence $\{a_n\}$ converges to $0\text{,}$ the distance between successive partial sums becomes as close to zero as we'd like, and thus the sequence of partial sums converges (even though we don't know the exact value to which it converges).

The preceding discussion has demonstrated the truth of the Alternating Series Test.

###### The Alternating Series Test

If $\{a_k\}$ is a sequence of positive terms that decreases to 0 as $k \to \infty\text{,}$ then the alternating series $\sum (-1)^{k}a_{k}$ converges.

Note that if the limit of the sequence $\{a_k\}$ is not 0, then the alternating series diverges.

###### Example7.40

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

2. $\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}2k}{k+5}$

3. $\displaystyle\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln(k)}$

1. $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+2}$ converges.

2. $\sum_{k=1}^{\infty} \frac{2(-1)^{k+1}k}{k+5}$ diverges.

3. $\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln(k)}$ converges.

Solution
1. Since $(k+1)^2 + 2 \gt k^2+2$ it follows that $\frac{1}{(k+1)^2+2} \lt \frac{1}{k^2+2}$ and the sequence $\frac{1}{k^2+2}$ decreases to 0. The Alternating Series Test shows then that the series $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+2}$ converges.

2. In this case we have that $\lim_{k \to \infty} \frac{2k}{k+5} = 2$ and so the sequence of $k$th terms does not even converge to 0. So the series $\sum_{k=1}^{\infty} \frac{2(-1)^{k+1}k}{k+5}$ diverges by the Divergence Test.

3. We know that the natural log is an increasing function, so $\frac{1}{\ln(k)}$ is a decreasing function. It is also true that $\lim_{k \to \infty} \frac{1}{\ln(k)} = 0$ and so the series $\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln(k)}$ converges by the Alternating Series Test.

### SubsectionSummary

• The ratio test may be used to test convergence by comparing to a geometric series. If the absolute value of the ratio of successive terms in the sequence is less than 1, the series converges. If this ratio is larger than 1, the series diverges. When the ratio is exactly one, the series may be convergent or divergent.

• An alternating series is a series whose terms alternate in sign. It has the form

\begin{equation*} \sum (-1)^ka_k \end{equation*}

where $a_k$ is a positive real number for each $k\text{.}$

• The sequence of partial sums of a convergent alternating series oscillates around the sum of the series if the sequence of $n$th terms converges to 0. That is why the Alternating Series Test shows that the alternating series $\sum_{k=1}^{\infty} (-1)^ka_k$ converges whenever the sequence $\{a_n\}$ of $n$th terms decreases to 0.

### SubsectionExercises

In this exercise we investigate the sequence $\left\{\frac{b^n}{n!}\right\}$ for any constant $b\text{.}$

1. Use the Ratio Test to determine if the series $\sum \frac{10^k}{k!}$ converges or diverges.

2. Now apply the Ratio Test to determine if the series $\sum \frac{b^k}{k!}$ converges for any constant $b\text{.}$

3. Use your result from (b) to decide whether the sequence $\left\{\frac{b^n}{n!}\right\}$ converges or diverges. If the sequence $\left\{\frac{b^n}{n!}\right\}$ converges, to what does it converge? Explain your reasoning.

There is a test for convergence similar to the Ratio Test called the Root Test. Suppose we have a series $\sum a_k$ of positive terms so that $a_n \to 0$ as $n \to \infty\text{.}$

1. Assume

\begin{equation*} \sqrt[n]{a_n} \to r \end{equation*}

as $n$ goes to infinity. Explain why this tells us that $a_n \approx r^n$ for large values of $n\text{.}$

2. Using the result of part (a), explain why $\sum a_k$ looks like a geometric series when $n$ is big. What is the ratio of the geometric series to which $\sum a_k$ is comparable?

3. Use what we know about geometric series to determine that values of $r$ so that $\sum a_k$ converges if $\sqrt[n]{a_n} \to r$ as $n \to \infty\text{.}$

In this exercise, we examine one of the conditions of the Alternating Series Test. Consider the alternating series

\begin{equation*} 1 - 1 + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{9} + \frac{1}{4} - \frac{1}{16} + \cdots\text{,} \end{equation*}

where the terms are selected alternately from the sequences $\left\{\frac{1}{n}\right\}$ and $\left\{-\frac{1}{n^2}\right\}\text{.}$

1. Explain why the $n$th term of the given series converges to 0 as $n$ goes to infinity.

2. Rewrite the given series by grouping terms in the following manner:

\begin{equation*} (1 - 1) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{9}\right) + \left(\frac{1}{4} - \frac{1}{16}\right) + \cdots\text{.} \end{equation*}

Use this regrouping to determine if the series converges or diverges.

3. Explain why the condition that the sequence $\{a_n\}$ decreases to a limit of 0 is included in the Alternating Series Test.