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Alternating and Absolute Convergence · Tutorial 651 of 998

Proof by Even and Odd Partial Sums

Use paired partial sums to prove convergence of an alternating series and locate its limit between consecutive even and odd partial sums.

Intermediate 10 min read

What You'll Learn

  • Define even and odd partial sums for an alternating series
  • Show that the even partial sums increase under the test’s hypotheses
  • Show that the odd partial sums decrease and remain above the even partial sums
  • Prove that both sequences of partial sums have the same limit
  • Derive the first-omitted-term error bounds from the partial-sum inequalities
  • Identify why alternating signs alone do not guarantee convergence

Why Split the Partial Sums by Parity?

An alternating series changes sign from one term to the next. That structure can make its partial sums easier to understand when we separate them into even-indexed and odd-indexed partial sums. Rather than trying to follow every step of one sequence of sums, we can study two sequences: one that moves upward and one that moves downward.

The argument proves the convergence conclusion of the alternating-series test. It also gives more than convergence: at every stage, the common limit lies between a pair of partial sums. Those inequalities provide bounds on the difference between the limit and either partial sum.

Setup and Notation

Consider the alternating series

$$ \sum_{n=1}^{\infty}(-1)^{n-1}a_n = a_1-a_2+a_3-a_4+\cdots, $$

where the terms satisfy

$$ a_n\geq 0,\qquad a_{n+1}\leq a_n,\qquad \lim_{n\to\infty}a_n=0. $$

Thus the magnitudes are nonnegative, nonincreasing, and tend to zero. For \(n\geq 1\), the \(n\)th partial sum is

$$ S_n=\sum_{j=1}^{n}(-1)^{j-1}a_j. $$

We also define \(S_0=0\), the sum with no terms. For each integer \(k\geq 0\), write

$$ E_k=S_{2k},\qquad O_k=S_{2k+1}. $$

Here \(E_k\) is an even partial sum and \(O_k\) is the following odd partial sum. In particular, \(E_0=0\) and \(O_0=a_1\). The definitions ensure that \(E_k\) and \(O_k\) are consecutive partial sums, so their difference is one term.

Definition: The even partial sums are \(E_k=S_{2k}\), and the odd partial sums are \(O_k=S_{2k+1}\), for \(k\geq 0\). We use this indexing so that both sequences begin at \(k=0\).

Worked Example: A Geometric Alternating Series

Worked Example: A Geometric Alternating Series

Take \(a_n=(1/2)^{n-1}\). These magnitudes are nonnegative, decrease, and tend to zero. The series is \(1-\frac12+\frac14-\frac18+\cdots\). Pairing successive terms gives

$$ E_k=\sum_{j=1}^{k}\left(\frac{1}{4^{j-1}}-\frac{1}{2\cdot 4^{j-1}}\right) =\sum_{j=1}^{k}\frac{1}{2\cdot 4^{j-1}} =\frac{2}{3}(1-4^{-k}). $$

The next odd partial sum adds \(a_{2k+1}=4^{-k}\), so

$$ O_k=E_k+4^{-k} =\frac{2}{3}+\frac{1}{3}4^{-k}. $$

As \(k\) tends to infinity, \(4^{-k}\) tends to zero. Both expressions therefore tend to \(2/3\). The inequalities are visible directly:

$$ E_k=\frac{2}{3}-\frac{2}{3}4^{-k} \leq \frac{2}{3} \leq \frac{2}{3}+\frac{1}{3}4^{-k}=O_k. $$

This verifies the common limit and the bracket for this example. For instance, at \(k=1\), \(E_1=1/2\) and \(O_1=3/4\), so \(1/2\leq 2/3\leq 3/4\), as required.

Why the Even Partial Sums Increase

Going from \(E_k\) to the next even partial sum \(E_{k+1}\) adds two terms, first positive and then negative:

$$ E_{k+1}-E_k =S_{2k+2}-S_{2k} =a_{2k+1}-a_{2k+2}. $$

Since the magnitudes are nonincreasing, \(a_{2k+1}\geq a_{2k+2}\). Hence \(E_{k+1}-E_k\geq 0\), and the even partial sums are nondecreasing.

The odd partial sums move in the other direction. From \(O_k=S_{2k+1}\) to \(O_{k+1}=S_{2k+3}\), the two added terms are negative and then positive:

$$ O_{k+1}-O_k =S_{2k+3}-S_{2k+1} =-a_{2k+2}+a_{2k+3}. $$

Again using the nonincreasing magnitudes, \(a_{2k+2}\geq a_{2k+3}\), so \(O_{k+1}-O_k\leq 0\). Therefore the odd partial sums are nonincreasing.

At each index \(k\), the difference between the odd and even partial sums is

$$ O_k-E_k=S_{2k+1}-S_{2k}=a_{2k+1}\geq 0. $$

Thus each even partial sum is no greater than the odd partial sum beside it. The even sequence rises, the odd sequence falls, and the distance between them is the next magnitude \(a_{2k+1}\).

The Convergence Proof

We can now prove that the two sequences of partial sums converge to the same real number. The argument uses the monotone convergence theorem for real sequences: a nondecreasing sequence bounded above converges, and a nonincreasing sequence bounded below converges.

Theorem: If \(a_n\geq 0\), \(a_{n+1}\leq a_n\), and \(a_n\to 0\), then the partial sums of \(\sum_{n=1}^{\infty}(-1)^{n-1}a_n\) converge.
1
Establish monotonicity.
The pairwise differences show \(E_{k+1}-E_k=a_{2k+1}-a_{2k+2}\geq 0\), while \(O_{k+1}-O_k=-a_{2k+2}+a_{2k+3}\leq 0\). Thus \(E_k\) is nondecreasing and \(O_k\) is nonincreasing.
2
Establish bounds.
Since \(E_k\leq O_k\) and \(O_k\leq O_0=a_1\), the sequence \(E_k\) is bounded above. Also \(E_k\geq E_0=0\) and \(E_k\leq O_k\), so \(O_k\) is bounded below.
3
Take the two limits.
By the monotone convergence theorem, there are real numbers \(E\) and \(O\) such that \(E_k\to E\) and \(O_k\to O\).
4
Show the limits agree.
The gap is \(O_k-E_k=a_{2k+1}\), which tends to zero. Therefore \(O-E=0\), so \(O=E\). Call their common value \(L\).
5
Conclude convergence of all partial sums.
The even partial sums \(S_{2k}=E_k\) and odd partial sums \(S_{2k+1}=O_k\) both tend to \(L\). These two subsequences together contain every partial sum, so \(S_n\to L\). By the definition of convergence of a series, the alternating series converges to \(L\).

The key point in the proof is not merely that each parity subsequence has a limit. They must have the same limit for the full sequence of partial sums to converge. The assumption \(a_n\to 0\) supplies exactly what is needed: the gap \(O_k-E_k=a_{2k+1}\) shrinks to zero.

Worked Example: The Alternating Harmonic Series

Worked Example: The Alternating Harmonic Series

For \(a_n=1/n\), the magnitudes are nonnegative, nonincreasing, and tend to zero, so the proof applies. The first even partial sums are

$$ E_1=S_2=1-\frac12=\frac12, \qquad E_2=S_4=1-\frac12+\frac13-\frac14=\frac{7}{12}. $$

The next one is

$$ E_3=S_6 =\frac{7}{12}+\frac15-\frac16 =\frac{7}{12}+\frac{1}{30} =\frac{37}{60}. $$

The corresponding odd partial sums are obtained by adding the next positive term:

$$ O_1=S_3=\frac12+\frac13=\frac56, \qquad O_2=S_5=\frac{7}{12}+\frac15=\frac{47}{60}, $$

and

$$ O_3=S_7=\frac{37}{60}+\frac17=\frac{319}{420}. $$

The brackets are in the correct order: \(E_3=259/420\), while \(O_3=319/420\). Thus the common limit \(L\) satisfies

$$ \frac{37}{60}\leq L\leq\frac{319}{420}, \qquad O_3-E_3=\frac{60}{420}=\frac17=a_7. $$

The calculation checks the predicted gap exactly. It illustrates how the proof brackets the limit, even when the limit itself has not been calculated.

The Limit Lies Between Consecutive Partial Sums

The proof also gives useful inequalities at every stage. Since \(E_k\leq O_k\), taking limits gives \(E_k\leq L\leq O_k\). Applying the same order to the next even partial sum gives \(E_{k+1}\leq L\leq O_k\). These statements yield bounds for the error after either partial sum.

$$ |L-S_{2k}|=L-E_k \leq O_k-E_k =a_{2k+1}, $$
$$ |L-S_{2k+1}|=O_k-L \leq O_k-E_{k+1} =a_{2k+2}. $$

The first equality in each line follows from the order of the terms: \(E_k\leq L\) and \(L\leq O_k\). The final equalities follow by subtracting the partial sums: \(O_k-E_k=a_{2k+1}\), and \(O_k-E_{k+1}=a_{2k+2}\). Thus, after an even partial sum, the error is at most the first omitted magnitude; after an odd partial sum, the same is true for the first omitted magnitude.

For the alternating harmonic example at \(k=3\), these inequalities give

$$ 0\leq L-S_6\leq a_7=\frac17, \qquad 0\leq S_7-L\leq a_8=\frac18. $$

These bounds are consequences of the same bracketing argument; they do not require a separate estimate of the unknown limit.

Worked Example: Equal Consecutive Magnitudes

Worked Example: Equal Consecutive Magnitudes

Consider magnitudes defined in pairs by \(a_{2j-1}=a_{2j}=2^{-(j-1)}\), for \(j\geq 1\). They are nonnegative and nonincreasing, and the values tend to zero as the pair index \(j\) increases. Each even partial sum cancels term by term:

$$ E_k=\sum_{j=1}^{k}(a_{2j-1}-a_{2j}) =\sum_{j=1}^{k}\left(2^{-(j-1)}-2^{-(j-1)}\right)=0. $$

The odd partial sum adds one more positive term after those cancellations:

$$ O_k=E_k+a_{2k+1}=0+2^{-k}=2^{-k}. $$

Therefore \(E_k\to 0\) and \(O_k\to 0\), and the common limit is \(L=0\). This is consistent with the actual partial sums: every even partial sum is zero, and the odd partial sums \(2^{-k}\) tend to zero. The example also shows that strict decrease is not needed; nonincreasing magnitudes suffice.

A Common Mistake: Alternation Alone Is Not Enough

Alternating signs do not, by themselves, guarantee convergence. Consider \(a_n=1\) for every \(n\). Then the terms alternate in sign, but their magnitudes do not tend to zero. The partial sums are

$$ S_1=1,\quad S_2=0,\quad S_3=1,\quad S_4=0,\quad \ldots $$

In the parity notation, \(E_k=0\) and \(O_k=1\) for every \(k\). The two sequences do not approach one another, so the partial sums have no common limit. The issue is precisely that their gap is \(O_k-E_k=a_{2k+1}=1\), not a quantity tending to zero.

Another error is to treat the two added terms in an even partial sum as if they had the same sign. The correct calculation is

$$ E_{k+1}-E_k =(+a_{2k+1})+(-a_{2k+2}) =a_{2k+1}-a_{2k+2}, $$

not \(a_{2k+1}+a_{2k+2}\). Keeping the alternating signs visible is essential to getting the correct monotonicity.

What the Proof Establishes

The even-and-odd partial-sum argument turns the hypotheses of the alternating-series test into a direct convergence proof. Pairing terms makes the even partial sums nondecreasing; the odd partial sums are nonincreasing; and their difference tends to zero. Monotone convergence gives limits for both subsequences, while the shrinking gap forces those limits to agree.

The proof does not calculate the common limit in every case. It does, however, show where that limit lies and bounds the error after either an even or an odd partial sum by the first omitted magnitude. The distinction matters: convergence is a statement that partial sums approach a limit, while finding a numerical value for that limit is a separate task.

Check Your Understanding

Use the definitions \(E_k=S_{2k}\) and \(O_k=S_{2k+1}\) to answer these questions.

  1. Compute \(E_{k+1}-E_k\) and explain why it is nonnegative under the hypotheses.
  2. Compute \(O_{k+1}-O_k\) and explain why it is nonpositive.
  3. Why does \(O_k-E_k\to 0\) imply that the limits of the even and odd partial sums agree?
  4. For \(k\geq 0\), express \(O_k-E_k\) as a single magnitude \(a_n\).
  5. State the error bounds for \(S_{2k}\) and \(S_{2k+1}\) in terms of the first omitted magnitude.
  6. What fails in the parity argument for the series \(1-1+1-1+\cdots\)?

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