Proof of the Alternating Series Test

The Proof Strategy: Separate the Partial Sums by Parity

The Alternating Series Test states that an alternating series converges when its nonnegative magnitudes are nonincreasing and tend to zero. The key to proving this is to separate the partial sums into those with an even number of terms and those with an odd number. Each subsequence moves in just one direction: the even partial sums increase, while the odd partial sums decrease.

These two subsequences do not merely converge independently. The difference between the odd and even partial sums with neighboring indices is exactly one term, whose magnitude tends to zero. Thus the two subsequences approach the same limit. Since every partial sum belongs to one of them, the full sequence of partial sums converges as well.

We use the positive-first form of the alternating series, \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\), with \(b_n\geq0\), \(b_{n+1}\leq b_n\), and \(b_n\to0\). Write \(S_0=0\) and \(S_N=\sum_{n=1}^{N}(-1)^{n-1}b_n\) for \(N\geq1\). For each \(k\geq1\), the even partial sum \(S_{2k}\) is followed by the odd partial sum \(S_{2k+1}\); the even and odd subsequences are \((S_{2k})\) and \((S_{2k-1})\), respectively.

Monotonicity of the Even and Odd Partial Sums

The change from one even partial sum to the next is a pair of terms. The change from one odd partial sum to the next is also a pair, but starts with a negative term. The hypothesis \(b_{n+1}\leq b_n\) gives the signs of both changes.

Proof. The change between consecutive even partial sums is $$ S_{2k}-S_{2k-2}=b_{2k-1}-b_{2k}\geq0, $$ because \(b_{2k-1}\geq b_{2k}\). Thus the even partial sums are nondecreasing. The change between consecutive odd partial sums is $$ S_{2k+1}-S_{2k-1}=-b_{2k}+b_{2k+1}\leq0, $$ because \(b_{2k+1}\leq b_{2k}\). Thus the odd partial sums are nonincreasing. Also, $$ S_{2k-1}-S_{2k}=b_{2k}\geq0. $$ Finally, \(b_{2k}\to0\) since \(b_n\to0\). These identities give all the stated inequalities and the separation limit. \(\square\)

The inequalities show why neither the even nor the odd partial sums need move toward the limit from the same direction: the even sums move upward, and the odd sums move downward. The distance between them shrinks to zero, so they close in on one another.

Proof of the Alternating Series Test

Proof. First consider the positive-first series, and use the partial sums \(S_N\) defined above. The even partial sums are nondecreasing by the lemma. They are bounded above: for every \(k\geq1\), $$ S_{2k}\leq S_{2k-1}\leq S_1=b_1, $$ because the odd partial sums are nonincreasing. The even partial sums are therefore a nondecreasing sequence bounded above, so they converge to some real number \(A\).

The odd partial sums are nonincreasing. They are bounded below by zero, since $$ S_{2k-1}\geq S_{2k}\geq S_0=0. $$ Here the first inequality follows from the lemma, and the second follows because the even partial sums are nondecreasing from \(S_0=0\). Thus the odd partial sums converge to some real number \(B\). By the separation identity in the lemma, $$ S_{2k-1}-S_{2k}=b_{2k}\longrightarrow0. $$ Since \(S_{2k}\to A\) and \(S_{2k-1}\to B\), taking limits in this difference gives \(B-A=0\). Hence \(A=B\).

It remains to show that convergence of the even and odd subsequences to the same limit implies convergence of the full sequence. Let \(\varepsilon>0\). There are indices beyond which both \(|S_{2k}-A|<\varepsilon\) and \(|S_{2k-1}-A|<\varepsilon\). Every sufficiently late partial sum is either an even partial sum or an odd partial sum, so every sufficiently late \(S_N\) satisfies \(|S_N-A|<\varepsilon\). Therefore \(S_N\to A\), which proves convergence of the series.

If the first sign is negative, its \(N\)th partial sum is the negative of the \(N\)th partial sum for the positive-first series with the same magnitudes. The negative of a convergent sequence converges, so the negative-first series converges as well. \(\square\)

The Two Subsequences Bracket the Sum

The proof also tells us where the limit lies relative to the partial sums. The even partial sums increase to the limit, while the odd partial sums decrease to it. This gives a useful structural conclusion independent of any particular formula for the sum.

Proof. The even partial sums are nondecreasing and converge to \(S\). Therefore each even partial sum is at most \(S\): \(S_{2k}\leq S\). The odd partial sums are nonincreasing and converge to \(S\), so each odd partial sum is at least \(S\): \(S\leq S_{2k-1}\). Combining these inequalities proves the claim. \(\square\)

This bracketing says, in particular, that consecutive partial sums on either side of an even index can give lower and upper bounds: \(S_{2k}\leq S\leq S_{2k-1}\). The width of this bracket is $$ S_{2k-1}-S_{2k}=b_{2k}. $$ The Alternating Series Test’s hypotheses ensure that this width tends to zero. The Remainder Bound for a Convergent Alternating Series, established in the previous tutorial, gives a separate bound on the error after a specified number of terms.

Worked Examples

What the Hypotheses Do—and Do Not—Say

The proof depends on two distinct properties. Nonincreasing magnitudes make the even partial sums increase and the odd partial sums decrease. The limit-zero condition makes the distance between those subsequences vanish. Without that last condition, monotonicity alone need not make the two subsequences converge to the same value.

The test is a sufficient condition, not a claim that every convergent alternating series must have nonincreasing magnitudes. If the magnitudes fail to be nonincreasing, this proof no longer establishes the required opposite monotonicity of the even and odd partial sums; the test then gives no conclusion. Also, convergence by this test does not by itself establish absolute convergence. That requires a separate examination of the series of magnitudes.