Alternating Series Error Bound

When a series converges, you usually can’t add up infinitely many terms, so you stop after a few and use the partial sum as an estimate. The natural question is: how good is that estimate? For alternating series there is a beautifully simple answer: the error is no bigger than the first term you left out. This is how calculators and AP questions justify statements like “this estimate is within $0.001$ of the true value.”

Key ideas

The setup

Suppose an alternating series

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

meets the conditions of the alternating series test: the $a_n$ are positive, decreasing, and $\displaystyle\lim_{n \to \infty} a_n = 0$. Let $S_n$ be the $n$th partial sum (the sum of the first $n$ terms).

The error bound

The error (or remainder) is the difference between the true sum and your estimate, $S - S_n$. The alternating series error bound says

$$|S - S_n| \le a_{n+1}$$

In words: the size of the error is at most the absolute value of the first omitted term.

Why it works

Each new term overshoots in the opposite direction, but by less than the term before. So the partial sums bounce back and forth around $S$, closing in, and $S$ is always trapped between two consecutive partial sums $S_n$ and $S_{n+1}$. The distance from $S_n$ to $S_{n+1}$ is exactly $a_{n+1}$, so $S$ can’t be farther than that from $S_n$.

The sum $S$ of $\sum (-1)^{n+1}/n^2$ is trapped between $S_4$ and $S_5$, so the error of $S_4$ is less than the gap $a_5 = 0.04$.

The sign of the error

Because $S$ lies between $S_n$ and $S_{n+1}$, the first omitted term also tells you which side of $S$ your estimate is on:

First omitted term is	Then $S_n$ is	Because
positive	an underestimate ($S_n \lt S$)	adding it would move the sum up toward $S$
negative	an overestimate ($S_n \gt S$)	adding it would move the sum down toward $S$

How many terms do you need?

To guarantee an error less than some tolerance $E$, find the first $n$ with $a_{n+1} \lt E$. Then $S_n$ (the first $n$ terms) is close enough. Be careful about whether the series starts at $n = 0$ or $n = 1$ when you count terms.

Taylor polynomials and AP justification

Many Taylor polynomials, such as those for $\sin x$, $\cos x$ and $e^{-x}$, give alternating series when you plug in a number. Then the error of the Taylor polynomial is at most the first omitted nonzero term. (For series that aren’t alternating, use the Lagrange error bound instead.)

On the AP exam, say why the bound applies: “The series is alternating, the terms decrease in absolute value, and they approach $0$, so by the alternating series error bound, the error is at most the absolute value of the first omitted term, which is …”

Worked examples

Example 1: Bounding the error of a partial sum

Estimate $S = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ using the first four terms. Bound the error, and say whether the estimate is too big or too small.

Solution. The terms $a_n = \dfrac{1}{n^2}$ are positive, decreasing, and approach $0$, so the alternating series error bound applies.

$$S_4 = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} = \frac{115}{144} \approx 0.799$$

The first omitted term is $+\dfrac{1}{25}$, so

$$|S - S_4| \le \frac{1}{25} = 0.04$$

The first omitted term is positive, so $S_4$ is an underestimate. (Check: the exact sum is $\dfrac{\pi^2}{12} \approx 0.822$. The actual error is about $0.024$, which is less than $0.04$, and $S_4$ is indeed too small.)

Example 2: How many terms?

How many terms of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}$ are needed to guarantee the partial sum is within $0.0005$ of the true sum?

Solution. The terms $\dfrac{1}{n^3}$ are positive, decreasing, and approach $0$. If you add $n$ terms, the error is at most $a_{n+1} = \dfrac{1}{(n+1)^3}$. You need

$$\frac{1}{(n+1)^3} \lt 0.0005 \quad\Rightarrow\quad (n+1)^3 \gt 2000$$

Since $12^3 = 1728$ and $13^3 = 2197$, you need $n + 1 \ge 13$, so $n \ge 12$. Twelve terms are enough.

Example 3: A Taylor polynomial for cos x

Use $1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!}$ to approximate $\cos 0.5$ (radians). Show the error is less than $0.0001$, and say whether the approximation is too big or too small.

Solution. The Maclaurin series for cosine at $x = 0.5$ is

$$\cos 0.5 = 1 - \frac{0.5^2}{2!} + \frac{0.5^4}{4!} - \frac{0.5^6}{6!} + \cdots$$

This is alternating, and its terms decrease in absolute value toward $0$. The approximation is

$$1 - 0.125 + 0.0026042 = 0.8776042 \approx 0.878$$

The first omitted term is $-\dfrac{0.5^6}{6!} = -\dfrac{0.015625}{720}$, so

$$|\text{error}| \le \frac{0.015625}{720} \approx 0.0000217 \lt 0.0001$$

The first omitted term is negative, so the approximation is an overestimate. (Check: $\cos 0.5 \approx 0.8775826$, slightly less.)

Example 4: Working from a given series

The Maclaurin series for a function $f$ is

$$f(x) = 1 - \frac{x}{3} + \frac{x^2}{5} - \frac{x^3}{7} + \cdots + \frac{(-1)^n x^n}{2n + 1} + \cdots$$

Use the first three terms to approximate $f\left(\tfrac{1}{2}\right)$, and show the approximation differs from $f\left(\tfrac{1}{2}\right)$ by less than $\tfrac{1}{50}$.

Solution. At $x = \tfrac{1}{2}$:

$$f\left(\tfrac{1}{2}\right) \approx 1 - \frac{1}{6} + \frac{1}{20} = \frac{53}{60} \approx 0.883$$

The series at $x = \tfrac{1}{2}$ is alternating, and its terms $\dfrac{(1/2)^n}{2n + 1}$ decrease (both the numerator shrinks and the denominator grows) and approach $0$. So the error is at most the first omitted term:

$$\left|\frac{(1/2)^3}{7}\right| = \frac{1}{56} \lt \frac{1}{50}$$

Common mistakes

Using the last term you kept instead of the first one you left out. The bound is the next term, $a_{n+1}$. If you add terms through $\dfrac{1}{16}$, the bound is $\dfrac{1}{25}$, not $\dfrac{1}{16}$.

Applying the bound when the conditions fail. The error bound needs an alternating series whose terms decrease in absolute value toward $0$. For a series with all positive terms (like $e^{0.3}$‘s Maclaurin series), it does not apply; use the Lagrange error bound.

Skipping zero terms incorrectly in Taylor polynomials. For $\cos x$, after the $x^4$ term the next nonzero term is the $x^6$ term. The “first omitted term” means the first omitted nonzero term.

Miscounting terms. If a series starts at $n = 0$, the first four terms are $n = 0, 1, 2, 3$, and the first omitted term is the $n = 4$ term. Write the terms out if you’re unsure.

Saying the error equals the bound. The bound is a guarantee: the error is at most $a_{n+1}$. In Example 1, the bound is $0.04$ but the actual error is about $0.024$.

Getting the over/under rule backwards. Ask: “Would adding the next term push my estimate up or down?” If up, the estimate is currently too small (an underestimate).

Practice

1. (Warm-up) Find $S_5$ for $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ and give an upper bound for the error.

Solution

$$S_5 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} = \frac{47}{60} \approx 0.783$$

The terms $\dfrac{1}{n}$ are positive, decreasing, and approach $0$, so $|S - S_5| \le a_6 = \dfrac{1}{6} \approx 0.167$. (The true sum is $\ln 2 \approx 0.693$.)

2. (Warm-up) Is $S_3$ an overestimate or an underestimate of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$? Give the error bound.

Solution

$S_3 = 1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} \approx 0.870$. The terms $\dfrac{1}{\sqrt{n}}$ decrease toward $0$, so the bound applies. The first omitted term is $-\dfrac{1}{\sqrt{4}} = -\dfrac{1}{2}$, which is negative, so $S_3$ is an overestimate, with error at most $\dfrac{1}{2}$.

3. (Warm-up) Use the terms through $n = 4$ of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$ to estimate the sum. Bound the error.

Solution

$$1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} = \frac{9}{24} = 0.375$$

The terms $\dfrac{1}{n!}$ are decreasing (from $n = 1$ on) and approach $0$. The first omitted term is the $n = 5$ term, $-\dfrac{1}{120}$, so the error is at most $\dfrac{1}{120} \approx 0.008$, and the estimate is an overestimate. (The sum is $e^{-1} \approx 0.368$.)

4. (Core) The series $1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \cdots = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n - 1}$ converges to $\dfrac{\pi}{4}$. How many terms guarantee an error less than $0.01$?

Solution

After $n$ terms, the error is at most $a_{n+1} = \dfrac{1}{2(n + 1) - 1} = \dfrac{1}{2n + 1}$.

$$\frac{1}{2n + 1} \lt 0.01 \quad\Rightarrow\quad 2n + 1 \gt 100 \quad\Rightarrow\quad n \gt 49.5$$

So $50$ terms are needed. (This series converges very slowly!)

5. (Core) Approximate $\sin 0.2$ (radians) using $x - \dfrac{x^3}{3!}$. Give an error bound and say whether the approximation is too big or too small.

Solution

$$0.2 - \frac{0.2^3}{6} = 0.2 - 0.0013333 = 0.1986667 \approx 0.199$$

The series for $\sin 0.2$ is alternating with terms decreasing to $0$. The first omitted term is $+\dfrac{0.2^5}{5!} = \dfrac{0.00032}{120} \approx 0.0000027$. So the error is at most about $0.0000027$, and since that term is positive, the approximation is an underestimate. (Check: $\sin 0.2 \approx 0.1986693$.)

6. (Core) Use the Maclaurin series $e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \cdots$ with three terms to estimate $e^{-0.3}$. Show that the error is less than $0.005$.

Solution

$$e^{-0.3} \approx 1 - 0.3 + \frac{0.09}{2} = 0.745$$

At $x = 0.3$ the series is alternating, and the terms $\dfrac{0.3^n}{n!}$ decrease toward $0$. The first omitted term is $-\dfrac{0.3^3}{3!} = -0.0045$, so

$$|\text{error}| \le 0.0045 \lt 0.005$$

(Check: $e^{-0.3} \approx 0.7408$; the actual error is about $0.0042$.)

7. (Core) A function has Maclaurin series $f(x) = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n \cdot 2^n}$.

• (a) Use the first three terms to approximate $f(1)$.
• (b) Show that this approximation differs from $f(1)$ by less than $0.02$.
• (c) Is the approximation too big or too small?

Solution

(a) At $x = 1$, the terms are $\dfrac{(-1)^{n+1}}{n \cdot 2^n}$:

$$f(1) \approx \frac{1}{2} - \frac{1}{8} + \frac{1}{24} = \frac{5}{12} \approx 0.417$$

(b) The series is alternating, and $n \cdot 2^n$ increases, so the terms decrease toward $0$. The first omitted term is $-\dfrac{1}{4 \cdot 16} = -\dfrac{1}{64}$, so the error is at most $\dfrac{1}{64} \approx 0.0156 \lt 0.02$.

(c) The first omitted term is negative, so $\dfrac{5}{12}$ is too big (an overestimate). (In fact $f(x) = \ln\left(1 + \tfrac{x}{2}\right)$, and $f(1) = \ln 1.5 \approx 0.405$.)

8. (Challenge) What is the smallest degree of a Maclaurin polynomial for $\cos x$ that is guaranteed by the alternating series error bound to approximate $\cos 1$ within $0.0001$? Give the approximation.

Solution

$$\cos 1 = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \frac{1}{8!} - \cdots$$

The terms $\dfrac{1}{(2k)!}$ decrease to $0$. Look for the first one below $0.0001$: $\dfrac{1}{6!} = \dfrac{1}{720} \approx 0.00139$ is too big, but $\dfrac{1}{8!} = \dfrac{1}{40320} \approx 0.0000248$ works. So stop just before the $x^8$ term: use the polynomial through $x^6$, which has degree 6.

$$\cos 1 \approx 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} \approx 0.540278$$

(Check: $\cos 1 \approx 0.540302$.)

9. (Challenge) Consider $S = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n}{3^n}$.

• (a) Show that the terms $a_n = \dfrac{n}{3^n}$ are decreasing.
• (b) Find $S_4$ and use the error bound to give an interval that must contain $S$.

Solution

(a) Compare consecutive terms:

$$\frac{a_{n+1}}{a_n} = \frac{n + 1}{3^{n+1}} \cdot \frac{3^n}{n} = \frac{n + 1}{3n} \le \frac{2}{3} \lt 1 \quad \text{for } n \ge 1$$

so each term is smaller than the one before. Also $a_n \to 0$, so the series converges by the alternating series test.

(b)

$$S_4 = \frac{1}{3} - \frac{2}{9} + \frac{3}{27} - \frac{4}{81} = \frac{14}{81} \approx 0.1728$$

The first omitted term is $+\dfrac{5}{243} \approx 0.0206$. $S$ lies between $S_4$ and $S_5 = S_4 + \dfrac{5}{243} = \dfrac{47}{243}$:

$$0.1728 \lt S \lt 0.1934$$

(The exact sum is $\dfrac{3}{16} = 0.1875$, which is inside.)