Alternating Series Error Bound

#10.10 Alternating Series Error Bound

Welcome to AP Calc 10.10! In this lesson, you’ll learn how to estimate the accuracy of a partial sum for an alternating series.

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#➕ Alternating Series Error Bound Theorem

The error bound theorem for an alternating series states that for a convergent alternating series, $\sum^\infty_{n=1}(-1)^n\cdot a_n$, we can estimate its true value by using an error bound. The error bound is defined as $a_i$. We can estimate the true value of the sum ($s$) with the following equation: $|s-s_{i-1}|\leq a_i$ where $i$ is the first omitted term of our estimation.

That’s a lot of information—now let’s break it down with an example!

#🧱 Breaking Down the Theorem

The first thing we need is a convergent, alternating series. Let’s use this one:

$$\sum_{n=1}^\infty \frac{(-1)^n\cdot n^2}{4^n}$$

We’re going to find the error bound for an estimation made with the first 3 terms. That means our bound is defined as $a_4$. Notice that we need to find up to the fourth term to find the error bound. This is because four is our first omitted term. For now, let’s calculate this fourth term.

$$a_4=\frac{(-1)^4\cdot 4^2}{4^4}=\frac{4^2}{4^4}=\frac{1}{4^2}=\boxed{\frac{1}{16}}$$

This is our error bound. That means that when we estimate the sum of the infinite series using the third term, our error won’t be greater than $\frac{1}{16}$ in either direction. Let’s finish this example by using the sum of the first three terms and the error bound to estimate what our infinite series is equal to.

$$\sum_{n=1}^3 \frac{(-1)^n\cdot n^2}{4^n}=-\frac{9}{64}$$

We can now say that the error bound is:

$$\left|s-\left(-\frac{9}{64}\right)\right|\leq \frac{1}{16}$$

We can simplify this like so:

$$-\frac{1}{16}\leq s+\frac{9}{64}\leq \frac{1}{16}\rightarrow -\frac{1}{16}-\frac{9}{64}\leq s\leq \frac{1}{16}-\frac{9}{64}\rightarrow \boxed{-\frac{13}{64}\leq s \leq -\frac{5}{64}}$$

This means that the true value of the infinite sum lies somewhere between about –0.203125 and –0.078125. You can also tell that this estimation isn’t very accurate, since the two bounds don’t share any digits. You can make more accurate estimations by calculating more terms.

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#📝 Alternating Series Error Bound Practice

Now it’s your turn to apply what you’ve learned!

#❓Alternating Series Error Bound Problems

Find the error bound of $\sum_{n=1}^5$ for the following infinite series and state which is a more accurate estimation.

1. $\sum \frac{(-1)^n}{n}$
2. $\sum \frac{(-1)^n(1+n^2)}{n^6+6}$

#💡 Alternating Series Error Bound Solutions

For each problem, first find $a_6$, the value of the first omitted term:

$$1. \ a_6=\frac{(-1)^6}{6}=\frac{1}{6}$$

$$2. \ a_6=\frac{(-1)^6(1+6^2)}{6^6+6}=\frac{1(1+36)}{46656+6}=\frac{37}{46662}$$

Then, find the value of the series up to five terms (review Riemann sums for help calculating summation!).

$$1. \ \sum_{n=1}^5 \frac{(-1)^n}{n}=\frac{-47}{60}\approx 0.783333$$

$$2. \ \sum_{n=1}^5 \frac{(-1)^n(1+n^2)}{n^6+6}\approx −0.225410$$

Next, set up your inequalities.

$$1. \ |s-s_5|\leq \frac{1}{6}$$

$$2. \ |s-s_6|\leq \frac{37}{46662}$$

Finally, simplify!

$$1. \ |s-s_5|\leq \frac{1}{6}\rightarrow -\frac{1}{6}\leq s-s_5\leq \frac{1}{6}\rightarrow -\frac{1}{6}\leq s-0.783333\leq \frac{1}{6}\rightarrow \boxed{0.616667\leq s\leq 0.95}$$

$$2. \ |s-s_5|\leq \frac{37}{46662}\rightarrow -\frac{37}{46662}\leq s-(−0.225410)\leq \frac{37}{46662}\rightarrow \\ -\frac{37}{46662}\leq s+0.225410\leq \frac{37}{46662}\rightarrow \boxed{-0.226203\leq s \leq -0.224617}$$

Now that we have both our answers, we can compare their accuracy. For the first problem, we see that the answers don’t share any decimal points—meaning it’s not very accurate. However, the second problem is accurate to two decimal places. Therefore, the second estimation is more accurate.

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#💫 Closing

As complicated as this might feel, this skill relies on techniques you’ve been learning for a long time! Make sure you keep practicing and you’re sure to ace any questions on this topic 💯