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Convergence Tests · Tutorial 833 of 999

Alternating Series Test and Monotone Terms

Learn when an alternating series converges by checking that its term magnitudes eventually decrease to zero.

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What You'll Learn

  • Recognize an alternating series from its sign pattern.
  • State the Alternating Series Test, including its eventual-monotonicity condition.
  • Prove why the test guarantees convergence.
  • Check decrease algebraically by comparing consecutive term magnitudes.
  • Distinguish a finite exception to monotonicity from failure of the test.
  • Identify when an alternating series fails the test because its terms do not approach zero.

Why Alternating Signs Matter

A series can behave differently when its terms alternate in sign than when all of its terms are positive. Positive and negative contributions can partially cancel, which makes some alternating series converge even when the series of term magnitudes does not.

To decide when that cancellation is enough, we need to examine the size of each term. The key conditions are that the magnitudes eventually decrease and approach zero. Both conditions matter: a decreasing sequence of magnitudes that settles above zero is not enough, and terms that approach zero need not decrease in the required way.

Central idea: For an alternating series, convergence is guaranteed if its term magnitudes are nonnegative, eventually nonincreasing, and tend to zero.

Alternating Series and Their Magnitudes

An alternating series has terms whose signs alternate between positive and negative. A standard form is

$$ \sum_{n=1}^{\infty}(-1)^{n-1}b_n = b_1-b_2+b_3-b_4+\cdots, $$

where each \(b_n\) is nonnegative. The numbers \(b_n\) are the term magnitudes, since the absolute value of the \(n\)th term is \(b_n\).

Definition: The sequence \((b_n)\) is eventually nonincreasing if there is an integer \(N\) such that \(b_{n+1}\leq b_n\) for every \(n\geq N\). It tends to zero if \(\lim_{n\to\infty}b_n=0\).

“Eventually” allows a finite number of early terms to behave differently. A finite number of terms cannot change whether the infinite tail converges; it only changes the sum by a finite amount. Thus, the magnitude sequence need not decrease starting at \(n=1\).

The Alternating Series Test

Theorem: Alternating Series Test. Suppose \(b_n\geq 0\) for all sufficiently large \(n\), \(b_n\) is eventually nonincreasing, and \(\lim_{n\to\infty}b_n=0\). Then the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\) converges.

The theorem guarantees convergence when its conditions hold. It does not say that these conditions are necessary for every alternating series to converge. If a condition fails, the test is inconclusive unless another fact—such as the terms failing to approach zero—settles the question.

Why the Test Works

First suppose the conditions hold from the beginning, so that \(b_n\geq 0\), \(b_{n+1}\leq b_n\), and \(b_n\to 0\) for every relevant index. Let \(S_n\) be the \(n\)th partial sum. The even partial sums group the terms into pairs:

$$ S_{2m}=(b_1-b_2)+(b_3-b_4)+\cdots+(b_{2m-1}-b_{2m}). $$

Each pair is nonnegative because \(b_{2j-1}\geq b_{2j}\). Also, adding the next pair cannot decrease an even partial sum, since \(b_{2m+1}-b_{2m+2}\geq 0\). Therefore, the even partial sums are nondecreasing. They are bounded above by \(b_1\): each finite sum of pairs is at most \(b_1\), as can be seen from

$$ S_{2m} =b_1-(b_2-b_3)-(b_4-b_5)-\cdots-(b_{2m-2}-b_{2m-1})-b_{2m} \leq b_1. $$

The odd partial sums are nonincreasing, because

$$ S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\leq 0. $$

They are bounded below by zero: an odd partial sum can be grouped as \(b_1-b_2+b_3-b_4+\cdots+b_{2m+1}\), and each completed pair is nonnegative, as is the final term. Thus the even partial sums and the odd partial sums each converge, by the monotone convergence property for bounded sequences.

Their limits agree. Indeed, the difference between consecutive odd and even partial sums is

$$ S_{2m+1}-S_{2m}=b_{2m+1}\longrightarrow 0. $$

So the even and odd partial sums approach the same limit. Every partial sum belongs to one of these two subsequences, which means the full sequence of partial sums converges. By the definition of series convergence, the alternating series converges.

If the conditions hold only from \(n=N\) onward, apply this argument to the alternating tail beginning at \(N\). The omitted initial terms form a finite sum, which does not affect convergence. This proves the eventual version of the test as well.

Checking Decrease Algebraically

For many sequences, the important step is to verify \(b_{n+1}\leq b_n\). One direct method is to calculate \(b_n-b_{n+1}\). If the denominators are positive, a positive numerator proves that the difference is positive and hence that the sequence decreases. A zero difference is also allowed: the test requires nonincreasing magnitudes, not strictly decreasing ones.

Worked Example: A Rational Magnitude Sequence

Test the convergence of

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

Here \(b_n=1/(3n+2)\). Its denominator is positive for \(n\geq 1\), so \(b_n>0\). Compare consecutive magnitudes:

$$ b_n-b_{n+1} =\frac{1}{3n+2}-\frac{1}{3(n+1)+2} =\frac{1}{3n+2}-\frac{1}{3n+5} =\frac{(3n+5)-(3n+2)}{(3n+2)(3n+5)} =\frac{3}{(3n+2)(3n+5)}>0. $$

Thus \(b_{n+1}<b_n\) for every \(n\geq 1\). Also, as \(n\) grows without bound, \(3n+2\) grows without bound, so

$$ \lim_{n\to\infty}b_n =\lim_{n\to\infty}\frac{1}{3n+2} =0. $$

The magnitudes are positive, decreasing, and tend to zero. The Alternating Series Test therefore proves that the series converges.

Check: At \(n=1\), \(b_1=1/5\), and at \(n=2\), \(b_2=1/8\). The calculation gives \(b_1-b_2=3/(5\cdot 8)=3/40>0\), consistent with the general result.

Worked Example: Establishing Decrease for a Less Obvious Sequence

Test the convergence of

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

Set \(b_n=n/(n^2+1)\). The numerator and denominator are positive for \(n\geq 1\), so \(b_n>0\). To check whether the magnitudes decrease, subtract consecutive terms and use a common denominator:

$$ \begin{aligned} b_n-b_{n+1} &=\frac{n}{n^2+1}-\frac{n+1}{(n+1)^2+1}\\ &=\frac{n\big((n+1)^2+1\big)-(n+1)(n^2+1)} {(n^2+1)\big((n+1)^2+1\big)}\\ &=\frac{n(n^2+2n+2)-(n+1)(n^2+1)} {(n^2+1)\big((n+1)^2+1\big)}\\ &=\frac{n^3+2n^2+2n-(n^3+n^2+n+1)} {(n^2+1)\big((n+1)^2+1\big)}\\ &=\frac{n^2+n-1}{(n^2+1)\big((n+1)^2+1\big)}. \end{aligned} $$

The denominator is positive. For \(n\geq 1\), the numerator satisfies \(n^2+n-1\geq 1+1-1=1>0\). Therefore \(b_n-b_{n+1}>0\), so the magnitudes decrease. To check the limit, divide the numerator and denominator of \(b_n\) by \(n^2\):

$$ b_n=\frac{n}{n^2+1} =\frac{1/n}{1+1/n^2}\longrightarrow 0. $$

All the conditions of the Alternating Series Test hold, so the series converges.

Check: At \(n=1\), \(b_1=1/2\), and at \(n=2\), \(b_2=2/5\). The formula for their difference gives

$$ b_1-b_2=\frac{1^2+1-1}{(1^2+1)(2^2+1)}=\frac{1}{10}, $$

which agrees with \(1/2-2/5=1/10\).

Eventual Decrease Is Enough

A few initial magnitudes may fail to decrease without invalidating the test. The relevant condition is that they are nonincreasing from some index onward.

Worked Example: A Finite Exception to Monotonicity

Define \(b_1=1/4\), and \(b_n=1/n\) for \(n\geq 2\). Consider the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\). Its first two magnitudes increase, since \(b_1=1/4\) and \(b_2=1/2\). But for every \(n\geq 2\),

$$ b_n-b_{n+1} =\frac{1}{n}-\frac{1}{n+1} =\frac{(n+1)-n}{n(n+1)} =\frac{1}{n(n+1)}>0. $$

So the magnitudes decrease from \(n=2\) onward. They also tend to zero, because \(b_n=1/n\) for all \(n\geq 2\). The Alternating Series Test applies to the tail, and adding the first term \(b_1=1/4\) does not change convergence. Therefore, the series converges.

Check: The initial increase is real: \(b_2-b_1=1/2-1/4=1/4>0\). The test still applies because its decrease condition holds eventually.

When a Condition Fails

A common error is to say that the alternating signs alone guarantee convergence. They do not. If the terms of a series converge to a nonzero value or fail to approach zero, the series diverges by the divergence test. In particular, the Alternating Series Test cannot establish convergence without the zero-limit condition.

Worked Example: Increasing Magnitudes and a Failed Zero Limit

Consider

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

The magnitudes are \(b_n=n/(n+1)\). Subtract consecutive terms:

$$ b_n-b_{n+1} =\frac{n}{n+1}-\frac{n+1}{n+2} =\frac{n(n+2)-(n+1)^2}{(n+1)(n+2)} =\frac{n^2+2n-(n^2+2n+1)}{(n+1)(n+2)} =-\frac{1}{(n+1)(n+2)}<0. $$

The negative difference means \(b_{n+1}>b_n\): the magnitudes increase. For instance, \(b_1=1/2\) and \(b_2=2/3\). Also,

$$ \lim_{n\to\infty}b_n =\lim_{n\to\infty}\frac{n}{n+1} =1. $$

The series terms therefore do not tend to zero: their magnitudes tend to \(1\). The series diverges by the divergence test. In this example the Alternating Series Test does not apply, and the nonzero term limit gives a separate reason for the divergence.

Common Pitfalls

  • Reading the sign of a difference backward. If \(b_n-b_{n+1}<0\), then \(b_{n+1}>b_n\), so the magnitudes increase, not decrease.
  • Checking only the first few values. Numerical examples can reveal a pattern, but a proof requires establishing the inequality for all sufficiently large \(n\).
  • Ignoring the zero limit. Decreasing magnitudes may approach a positive number. The Alternating Series Test requires their limit to be zero.
  • Demanding decrease from the first term. A finite number of exceptions is permitted; the magnitudes only need to be nonincreasing eventually.
  • Treating a failed test as proof of divergence. If the magnitudes are not eventually nonincreasing, this test gives no conclusion by itself. Check whether the terms fail to approach zero or consider another appropriate argument.

Why the Conditions Work Together

Alternating signs create cancellation, but the cancellation is useful in a controlled way only when successive magnitudes do not grow and the contributions become arbitrarily small. Eventual decrease ensures that, after pairing consecutive terms, the partial sums move in an orderly direction. The zero limit ensures that the even and odd partial sums approach the same value. The proof above shows how these facts combine to guarantee convergence.

When applying the test, state each condition explicitly: identify the magnitude \(b_n\), establish \(b_n\geq 0\), verify eventual nonincrease, and calculate \(\lim_{n\to\infty}b_n\). Then make the convergence conclusion. This separates the algebraic work from the final justification and makes it easier to catch a reversed inequality or a missing limit condition.

Check Your Understanding

For each question, identify the magnitude sequence and state which condition or conclusion is relevant.

  1. State the three conditions in the Alternating Series Test.
  2. For \(b_n=1/(2n+1)\), calculate \(b_n-b_{n+1}\) and determine whether the magnitudes decrease.
  3. If \(b_n-b_{n+1}<0\), what does that tell you about the direction of change in the magnitudes?
  4. Can a finite increase among the first few magnitudes prevent the Alternating Series Test from applying? Explain.
  5. For \(b_n=n/(n+1)\), what is the limit of the magnitudes, and what does that imply about the alternating series?

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