The Alternating-Series Remainder Bound

Why a Remainder Bound Matters

An infinite series is often represented in practice by a finite sum. The difference between the series’ sum and that finite sum is its remainder. Knowing that the series converges does not, by itself, tell us how large this difference can be.

For an alternating series whose term magnitudes decrease to zero, the remainder has a particularly useful bound: its magnitude is no greater than the magnitude of the first term left out. This gives a direct way to assess the error made by stopping after a specified number of terms. The bound depends on the alternating-series conditions; it is not a general rule for all convergent series.

The Remainder and Its Bound

Consider an alternating series written as

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

where $b_n\ge 0$. Its $N$th partial sum is

S_N=\sum_{n=1}^{N}(-1)^{n-1}b_n.

When the series converges to a sum $S$, the remainder after $N$ terms is the difference between the full sum and that partial sum.

$$ R_N=S-S_N. $$

Definition: For a convergent series, the remainder after $N$ terms is $R_N=S-S_N$. For an alternating series with $b_n\ge 0$, $b_{N+1}$ is the magnitude of the first omitted term.

The alternating-series remainder bound applies when the term magnitudes are nonincreasing and approach zero:

b_{n+1}\le b_n\quad\text{for all }n, \qquad \lim_{n\to\infty}b_n=0.

Theorem: Alternating-series remainder bound. If $b_n\ge 0$, $b_{n+1}\le b_n$ for all $n$, and $\lim_{n\to\infty}b_n=0$, then the alternating series converges. If its sum is $S$, then for every $N\ge 1$, $$ |S-S_N|=|R_N|\le b_{N+1}. $$

The upper bound uses $b_{N+1}$, not $b_N$: the term indexed $N$ is already included in $S_N$. The next term is the first one omitted.

Why the Bound Holds

The alternating-series test guarantees convergence under the stated conditions, so the infinite tail after $S_N$ has a well-defined sum. To see why its magnitude is at most $b_{N+1}$, first examine finite portions of that tail. After factoring out the sign of its first term, such a portion has the form

T_m=b_{N+1}-b_{N+2}+b_{N+3}-\cdots+(-1)^{m-1}b_{N+m}.

We show that every $T_m$ lies between $0$ and $b_{N+1}$. For an even number $m=2r$ of terms, group consecutive terms from the start to see that

T_{2r} =(b_{N+1}-b_{N+2})+\cdots+ (b_{N+2r-1}-b_{N+2r})\ge 0,

because the magnitudes are nonincreasing. Grouping the same terms in a different way gives the upper bound:

T_{2r} =b_{N+1}-(b_{N+2}-b_{N+3})-\cdots -(b_{N+2r-2}-b_{N+2r-1})-b_{N+2r} \le b_{N+1}.

For an odd number $m=2r+1$ of terms, grouping from the start gives

T_{2r+1} =(b_{N+1}-b_{N+2})+\cdots+ (b_{N+2r-1}-b_{N+2r})+b_{N+2r+1}\ge 0.

Grouping the odd tail after its first term instead gives

T_{2r+1} =b_{N+1}-(b_{N+2}-b_{N+3})-\cdots -(b_{N+2r}-b_{N+2r+1}) \le b_{N+1}.

Thus $0\le T_m\le b_{N+1}$ for every finite $m$. As $m$ increases, these finite tails approach the infinite tail, because the original series converges. The limit therefore also lies between $0$ and $b_{N+1}$. Multiplying a tail by $-1$, if needed, does not change its absolute value. Hence the magnitude of the remainder is at most $b_{N+1}$, as claimed.

Worked Examples

Worked Example: The Alternating Harmonic Series After Five Terms

Consider $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}$. Determine the remainder bound after $N=5$ terms.

Here $b_n=1/n$, which is positive. Also, $b_{n+1}=1/(n+1)\le 1/n=b_n$, and $\lim_{n\to\infty}1/n=0$. The theorem therefore applies.

Compute the fifth partial sum:

S_5=1-\frac12+\frac13-\frac14+\frac15 =\frac{60-30+20-15+12}{60} =\frac{47}{60}.

The first omitted term has index $6$, so its magnitude is $b_6=1/6$. Therefore

|S-S_5|\le b_6=\frac16.

Check: the calculation uses five terms, with indices $1$ through $5$; index $6$ is indeed the first omitted index. The common denominator calculation gives $47/60$, and the theorem gives the stated bound $1/6$.

Worked Example: An Alternating Series of Reciprocal Squares

Consider $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n^2}$. Find the remainder bound after $N=3$ terms.

In this case $b_n=1/n^2>0$. Since $n+1>n>0$, we have $1/(n+1)^2<1/n^2$; and $1/n^2$ tends to zero. Thus the theorem applies.

The partial sum is

S_3=1-\frac14+\frac19 =\frac{36-9+4}{36} =\frac{31}{36}.

The first omitted term is indexed by $4$, and its magnitude is $b_4=1/4^2=1/16$. It follows that

|S-S_3|\le\frac{1}{16}.

Check: the three included terms have indices $1,2,3$, and the next index is $4$. The numerator $36-9+4=31$, so the computed partial sum and the first-omitted-term bound are consistent.

Worked Example: Alternating Odd Reciprocals After Four Terms

Consider $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2n-1}$. Determine the remainder bound after $N=4$ terms.

Here $b_n=1/(2n-1)$ is positive. As $n$ increases, the positive denominator increases, so $b_n$ decreases. Also, $1/(2n-1)$ tends to zero. Both conditions are satisfied.

The fourth partial sum is

S_4=1-\frac13+\frac15-\frac17 =\frac{105-35+21-15}{105} =\frac{76}{105}.

The first omitted term has index $5$, so its magnitude is

b_5=\frac{1}{2(5)-1}=\frac19.

Consequently,

|S-S_4|\le\frac19.

Check: substituting $n=5$ into $1/(2n-1)$ gives $1/(10-1)=1/9$, and the four terms in $S_4$ combine to $76/105$.

A Common Indexing Mistake

A frequent error is to use the term just before the first omitted term, or to skip ahead to a smaller term. For example, after five terms of the alternating harmonic series, the term of magnitude $1/5$ has already been included, and the next term has magnitude $1/6$. The theorem’s bound is therefore $1/6$. Claiming the stronger bound $1/7$ is not justified by this theorem: the first omitted term is not $1/7$.

N=5 \quad\Longrightarrow\quad \text{first omitted index}=N+1=6 \quad\Longrightarrow\quad |R_5|\le b_6=\frac16.

Another important limitation is that the conditions must hold for the term magnitudes in the tail under consideration. Alternation alone is not enough. If the magnitudes do not decrease to zero, this remainder theorem does not apply, even if the signs alternate.

When This Result Is Useful

The remainder bound translates the behavior of an infinite tail into information about one explicit term. Once the conditions are checked, the first omitted term supplies a guaranteed upper bound for the error after $N$ terms. This is why it is important to keep the index precise: the partial sum determines exactly where the tail begins.

The bound need not give the exact error. It is a guarantee, not an equality in general. Nor can it be used without checking the hypotheses. A sound application follows a short sequence: identify the term magnitudes, verify that they are nonincreasing and tend to zero, calculate the partial sum if needed, and use the magnitude at index $N+1$.

Check the conditions.
Confirm that $b_n\ge 0$, $b_{n+1}\le b_n$, and $b_n\to0$.

Locate the tail.
After $N$ included terms, the first omitted term has index $N+1$.

Apply the bound.
Use $|S-S_N|\le b_{N+1}$.

Check Your Understanding

Use the theorem’s conditions and indexing carefully in each question.

State the three conditions on $b_n$ required for the alternating-series remainder bound.
For a partial sum ending at index $N=8$, which term magnitude bounds the remainder?
For $\sum_{n=1}^{\infty}(-1)^{n-1}/(n+2)$, what is the first omitted term after six terms, and what is the resulting bound?
Why does knowing that the signs alternate, without checking the magnitudes, fail to justify the remainder bound?
After four terms of $\sum_{n=1}^{\infty}(-1)^{n-1}/n^2$, is the theorem’s bound $1/16$, $1/25$, or $1/36$? Explain the index used.