Remainders for alternating series

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\,d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}\hspace {0in}} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \DeclareMathOperator {\proj }{\mathbf {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

There is a nice result for approximating the remainder of convergent alternating series.

Introduction

Recall that if $\{a_n\}_{n=n_0}$ is a sequence of positive terms, we say the series $\sum _{k=n_0}^{\infty } (-1)^k a_k$ and $\sum _{k=n_0}^{\infty } (-1)^{k+1} a_k$ are alternating series, and we have a nice result to test these for convergence.

Alternating Series Test Let $\{a_n\}_{n=n_0}$ be a sequence. If

$a_n \geq > 0$ eventually,
$a_{n+1} \leq a_n$ eventually, and
$\lim _{n \to \infty } a_n=0$ ,

then, the alternating series $\sum _{k=n_0}^\infty (-1)^{k}a_k$ converges.

Note that this test gives us a way to determine that many alternating series must converge, but it does not give us information about their corresponding values. If we want to approximate such series, we must study their remainders.

Remainders for Alternating Series

As usual we must establish that a series converges first before we begin to think about remainders. Once we have established that an alternating series $\sum _{k=1}^{\infty } (-1)^k a_k$ converges, we have the usual decomposition.

As before, $s_n$ is the approximate value of the infinite series and $r_n$ is the error made when using this approximation. While we cannot find an explicit formula for $s_n$ , we have a good way to establish bounds on the error made when approximating $\sum _{k=1}^{\infty } (-1)^k a_k$ by the finite sum $s_n= \sum _{k=1}^{n} (-1)^k a_k$ . Let’s explore this result pictorially for a general alternating series.

Let’s begin with a convergent alternating series $\sum _{k=0}^{\infty } (-1)^k a_k$ for which the alternating series test applies. For the sake of argument, we make the following conventions to begin the example.

$a_n > 0$ for every $n \geq 0$ .
$\{a_n\}_{n=0}$ is strictly decreasing immediately; that is $a_{n+1} < a_n$ for every $n \geq 0$ .
$\lim _{n\to \infty } a_n= 0$ (not really an assumption since $\sum _{k=0}^{\infty } (-1)^k a_k$ converges).
The value of the convergent series $\sum _{k=0}^{\infty } (-1)^k a_k$ is the number $S$ .

Let’s plot the terms of two sequences : $\{a_n\}_{n=0}$ , which consists of positive terms and the sequence of partial sums $\{s_n\}_{n=0}$ for the alternating series $\sum _{k=0}^{\infty } (-1)^k a_k$ . We first start with the plot of $\{a_n\}_{n=0}$ .

We now plot the terms in the series $\sum _{k=0}^{\infty } (-1)^ka _k$ . Note that when written out, the sum in question is

$a_0-a_1+a_2+a_3-\ldots$ That is, we alternate between adding and subtracting terms in the sequence $\{a_n\}_{n=0}$ .

Note that in the above picture, the term $r_n$ corresponding to the error made when we use $s_n$ to approximate the infinite series is the difference between the actual value of the series, and $s_n$ .

We now note the following. For odd indices $n$ , we have (from the picture) the relationship below. $s_1 > s_3 > s_5 > s_7 > \dots > S,$ By the same logic about the relative sizes of the terms, we have for even $n$ that $s_2 < s_4 < s_6 < s_8 < \dots < S.$

Now, we can think of the remainder $r_n$ as the difference between $S$ and $s_n$ at any stage, or that

When $n$ is even, $s_n > S$ . But then $n+1$ is odd, meaning that $s_{n+1} < S$ . To get from $s_n$ to $s_{n+1}$ , we subtract $s_n - a_{n+1}$ . But this subtraction takes us from a number greater than $S$ to a number less than $S$ ‘. In other words, the distance between $S$ and $s_n$ is less than $a_{n+1}$ . In symbols, $\vert S - s_n \vert < a_{n+1}.$ Similarly, if $n$ is odd, then $n+1$ is even, so that $s_n < S$ and $s_{n+1} > S$ . Again, since $s_{n+1} = s_n + a_{n+1}$ , the distance between $S$ and $s_n$ is less that $a_{n+1}$ . But this distance is precisely the absolute value of $r_n$ .

Of course, if $a_1$ is negative, we have the same behavior, except the odd partial sums are increasing and the even ones are decreasing. Try it out with an example if you are skeptical! We can now state the general result for approximating alternating series.

Alternating Series Remainder Estimates Let $\{a_n\}_{n=n_0}$ be a sequence. If

$a_n \geq 0$ ,
$a_{n+1} \leq a_n$ , and
$\lim _{n \to \infty } a_n=0$ ,

then, we have the following estimate for the remainder.

$\big | r_n \big | \leq a_{n+1} \qquad \textrm { for all } n \geq n_0,$ where $r_n = \sum _{k=n+1}^{\infty } a_k$ .

Note that while the actual alternating series test requires that the terms $a_k$ in the series $\sum _{k=n_0} (-1)^k a_k$ or $\sum _{k=n_0} (-1)^{k+1} a_k$ eventually be positive and decreasing, the remainder results require this for all terms; that is $a_n$ must be positive and $a_{n+1} \geq a_n$ for all $n \geq n_0$ . If this is not the case, care must be taken when constructing the estimates. Details are left for the curious reader to ponder.

In order to gain some practice, let’s work an example.

Consider the series $\sum _{k=1}^\infty \frac {(-1)^k}{k^2}$ . Give an upper bound for the absolute value of the error involved with using $s_8$ in place of the exact value of the series, which we call $S$ .

First, notice that the series is a convergent alternating series - so it makes sense to even ask the question in the first place!

Second, our error estimate is simply the next term of the series. In particular,

$\begin{align*} \vert r_8 \vert &< a_{\answer[given]{9}}\\ &= \answer[given]{1/81}. \end{align*}$

Simple enough! Let’s see if our other typical question presents any additional trouble.

How many terms of the series $\sum _{k=1}^{\infty }\frac {(-1)^k}{k^2}$ do we need to add in order to guarantee that the estimate $s_n$ is within $0.01$ of the actual sum $S$ ?

Here we are looking for a value of $n$ after which we are guaranteed small enough error. Again, we will use the Alternating Series Test to answer this question.

We know that $\vert r_n \vert \leq a_{\answer [given]{n+1}},$ so if we can force our upper bound to be less than $0.01$ , then the size of the remainder can be no larger. Now

$\begin{align*} a_{\answer[given]{n+1}} &\leq 0.01 \\ \frac{1}{ \answer[given]{(n+1)^2}} & \leq \frac{1}{100} \end{align*}$ is satisfied for $100 \leq \answer [given]{(n+1)^2}.$ Taking the square root of both sides, we see that we need $10 \leq \answer [given]{n+1}$ or $n \geq \answer [given]{9}.$ In other words, when $n \geq \answer [given]{9}$ , we have $\vert r_n \vert \leq a_{n+1} \leq 0.01.$

Note that the above guarantees that for any integer $N>n_0$ , the term $s_N$ will approximate the series $\sum _{k=1}^{\infty } a_k$ with an error of magnitude no greater than $a_{n+1}$ . Since we are now working with series which have both positive and negative terms, we have switched to talking about the size of the error rather than its actual value. The remainder itself could be positive or negative, depending on our series and the chosen value of $n$ . The size, or absolute value, of the remainder gives us the information we usually want in practice – how far are we from the sum? But if an application problem is asking you whether the partial sum is above or below the sum, you can now answer that question as well!

Finally, let’s consider one more problem.

Determine if there is a value for $N$ so $\sum _{k=1}^{N} (-1)^k \cdot \frac {k+2}{2k+1}$ is within $.01$ of the value of $\sum _{k=1}^{N} (-1)^k \cdot \frac {k+2}{2k+1}$ . If there is such a value, give one possibility for it.

There is no such value for $N$ since the series $\sum _{k=1}^{N} (-1)^k \cdot \frac {k+2}{2k+1}$ diverges. There is such a value for $N$ .

Note that before we try to approximate the value of a series, we must first determine that it converges (so it actually has a value). Here, note that $\lim _{n \to \infty } \frac {n+2}{2n+1} = \frac {1}{2}$ , so $\lim _{n \to \infty } (-1)^n \cdot \frac {n+2}{2n+1}$ does not exist, and hence $\sum _{k=1}^{N} (-1)^k \cdot \frac {k+2}{2k+1}$ diverges by the divergence test.

Summary

When $\{a_n\}_{n = n_0}$ is a decreasing sequence of positive term, we will approximate $\sum _{k=n_0}^{\infty } a_k$ by the finite sum $s_n =\sum _{k=n_0}^{n} a_k$ . Generally, adding more and more terms of a convergent series should generally get you closer to the actual sum! Indeed, we have a nice bound for the remainder:

$\textrm {magnitude of error} = |r_n| \leq a_{n+1}.$

We can use to approximate the series to any degree of accuracy as we want!