Infinite Sequences, Series, and Error

Motivation

You need Chapter 5's finite sums, elementary inequalities and powers. You should learn to distinguish a sequence from a series, decide what convergence asserts, calculate geometric and telescoping series through their partial sums, and justify a finite approximation with an error bound. Infinite notation does not grant permission to treat an endless list like a finite expression.

An instrument may record one value at each integer time. An echo model may add successively weaker signals. In the first problem we ask what happens to the individual values; in the second we ask what happens to their running total. These are different sequences and can behave differently.

6.1 Sequences and indexing

An infinite real sequence in this chapter is a function from an integer tail $\{n_0,n_0+1,n_0+2,\ldots\}$ to $\mathbb R$, where $n_0$ is a fixed integer. The index set therefore contains every sufficiently large integer and has no last index; the usual choices are $n_0=1$ or $n_0=0$. Its values are written $a_n$. The notation $(a_n)$ denotes the whole sequence, not one unspecified term.

An explicit formula computes a term directly. A recurrence specifies a starting value and a rule for computing later terms. Both must include an index convention. For example,

$$a_n=3+2n\quad(n\ge0)$$

starts $3,5,7,\ldots$, whereas the same formula with $n\ge1$ starts $5,7,9,\ldots$. For a recurrence $a_{n+1}=2a_n$, no unique sequence exists until $a_0$ or another appropriate initial value is given.

An arithmetic sequence has a fixed difference; a geometric sequence has a fixed multiplier. If the first term is indexed zero, these have forms $a_n=a_0+nd$ and $a_n=a_0r^n$. When $r=0$, use $a_0$ as the given starting value and set all later terms to zero; no general convention for $0^0$ is needed.

Worked example

Worked example 1: explicit and recursive descriptions. Let $u_0=7$ and $u_{n+1}=(u_n+3)/2$ for $n\ge0$. Subtract the equilibrium value three:

$$u_{n+1}-3=\frac12(u_n-3).$$

Therefore $u_n-3=4(1/2)^n$, so $u_n=3+4/2^n$. This is verified by substitution into the recurrence and the initial condition; Chapter 7 formalizes the induction behind such verification. The terms decrease towards three and never equal three at a finite index.

6.2 What a limit means

Definition

A sequence converges to a real number $L$, written $a_n\to L$, when

for every $\varepsilon>0$, there is an integer $N$ such that $|a_n-L|<\varepsilon$ for every allowed index $n\ge N$.

The positive number $\varepsilon$ is an error tolerance. The index $N$ can depend on the tolerance but must control the whole tail, not merely a specially chosen term. A sequence that does not converge to any finite real number is divergent. A sequence can diverge by growing without bound or by oscillating, among other possibilities.

Worked example

Worked example 2: an explicit tolerance. For $a_n=2+3/n$, $n\ge1$, the distance from two is $3/n$. Given $\varepsilon>0$, choose an integer $N>3/\varepsilon$. Then $n\ge N$ implies $|a_n-2|=3/n\le3/N<\varepsilon$. Thus $a_n\to2$. For error strictly below $0.01$, $N=301$ works. Choosing $N=300$ would give equality, not a strict inequality, at the first index.

A limit is unique. If $L\ne M$ were both limits, take $\varepsilon=|L-M|/3$. Far enough along the sequence both $|a_n-L|$ and $|a_n-M|$ would be less than $\varepsilon$. The triangle inequality would give $|L-M|<2|L-M|/3$, a contradiction.

Convergent sequences are bounded. To see this, choose a tail with $|a_n-L|<1$, so its terms have magnitude below $|L|+1$. There are only finitely many earlier terms; taking the maximum of their magnitudes and $|L|+1$ bounds the whole sequence. The converse is false: $(-1)^n$ is bounded but cannot converge because its even and odd terms remain two units apart.

A sequence is nondecreasing if $a_{n+1}\ge a_n$ and nonincreasing if $a_{n+1}\le a_n$ for every relevant $n$. A monotone sequence can be constant. Neither monotonicity alone nor boundedness alone guarantees convergence. For example, $a_n=n$ is increasing but unbounded.

6.3 Elementary limit tools and their hypotheses

Proposition

If $a_n\to A$ and $b_n\to B$, then for fixed real $c$,

$$a_n+b_n\to A+B,\quad ca_n\to cA,\quad a_nb_n\to AB.$$

If in addition $B\ne0$, then $b_n\ne0$ for all sufficiently large $n$, and $a_n/b_n\to A/B$ on that tail. A finite number of omitted or changed initial terms does not affect a limit.

Proof

Here is the reasoning behind the algebra. For sums, the error is at most $|a_n-A|+|b_n-B|$; make both smaller than half a desired tolerance. For products, write

$$a_nb_n-AB=a_n(b_n-B)+B(a_n-A).$$

The sequence $a_n$ is bounded, so sufficiently small errors in each factor make the product error small. For quotients, $b_n\to B\ne0$ implies eventually $|b_n|>|B|/2$, and

$$\left|\frac1{b_n}-\frac1B\right| =\frac{|b_n-B|}{|b_nB|}\le\frac{2|b_n-B|}{|B|^2}\to0.$$

Then apply the product rule. Dividing by a sequence tending to zero is not covered by this rule.

Complete tolerance choices for the algebraic limit rules

Let $a_n\to A$ and $b_n\to B$ be real sequences indexed by the integers $n\ge n_0$, and fix $\varepsilon>0$. All thresholds below may be enlarged to be at least $n_0$. For the sum, choose thresholds for $|a_n-A|<\varepsilon/2$ and $|b_n-B|<\varepsilon/2$, then take their maximum. The triangle inequality gives $|(a_n+b_n)-(A+B)|<\varepsilon$. For fixed real $c$, if $c=0$ the scalar-multiple result is immediate; otherwise choose $|a_n-A|<\varepsilon/|c|$, giving $|ca_n-cA|<\varepsilon$. This also proves the difference rule by setting $c=-1$ and using the sum rule.

For the product, boundedness proved in Section 6.2 supplies a real $M>0$ with $|a_n|\le M$ for every index. Choose a common tail on which

$$|b_n-B|<\frac{\varepsilon}{2M},\qquad |a_n-A|<\frac{\varepsilon}{2(|B|+1)}.$$

The displayed product decomposition in Section 6.3 then yields

$$|a_nb_n-AB|\le M|b_n-B|+|B|\,|a_n-A|<\frac{\varepsilon}{2}+\frac{\varepsilon|B|}{2(|B|+1)}<\varepsilon.$$

For reciprocals assume $B\ne0$. Choose a common tail with $|b_n-B|<|B|/2$ and $|b_n-B|<\varepsilon|B|^2/2$. Since $|B|\le|B-b_n|+|b_n|$, the first inequality gives $|b_n|>|B|/2>0$. Therefore

$$\left|\frac1{b_n}-\frac1B\right|=\frac{|b_n-B|}{|b_n|\,|B|}<\frac{2|b_n-B|}{|B|^2}<\varepsilon.$$

Thus $1/b_n\to1/B$ on the nonzero tail, and the proved product rule gives $a_n/b_n\to A/B$. A finite change or omission affects no sufficiently late tail, so none of these conclusions depends on such initial terms. Only real arithmetic, the triangle inequality, the definition of convergence and the previously proved boundedness theorem are used; no limit rule is assumed in its own proof.

If $\ell_n\le a_n\le u_n$ eventually, and both $\ell_n,u_n\to L$, then $a_n\to L$. Indeed, for a given tolerance the lower and upper bounds eventually lie between $L-\varepsilon$ and $L+\varepsilon$, trapping $a_n$ there. This is the squeeze principle.

Worked example

Worked example 3: normalize a rational expression. For $n\ge1$,

$$\frac{5n-2}{2n+7}=\frac{5-2/n}{2+7/n}\longrightarrow\frac52.$$

The denominator tends to two, not zero, so the quotient rule applies. In contrast $n/n$ cannot be handled by separately assigning a finite limit to its numerator and denominator; neither sequence converges to a finite real number. It is simply one at every index.

For real $r$ with $|r|<1$, $r^n\to0$. For $0<q=|r|<1$, put $h=q^{-1}-1>0$. The elementary inequality $(1+h)^n\ge1+nh$ gives

$$|r^n|=q^n\le\frac1{1+nh}\to0.$$

The inequality follows by multiplying $1+nh$ by $1+h$ and retaining the extra nonnegative $nh^2$ at each step, starting at $n=1$. For $r=0$, all positive-index powers are zero. At $r=1$ the sequence is one; at $r=-1$ it oscillates; for $|r|>1$ its magnitude grows without bound. These boundary cases must not be hidden in the formula for $|r|<1$.

A further foundational fact is monotone bounded convergence: every nondecreasing sequence bounded above converges, and every nonincreasing sequence bounded below converges. For the first statement, let $L$ be the least upper bound of its terms, using completeness of the real numbers. Given $\varepsilon>0$, $L-\varepsilon$ cannot be an upper bound, so some term exceeds it. Monotonicity makes every later term lie between that term and $L$, hence within $\varepsilon$ of $L$. Negating a sequence proves the decreasing case. This theorem establishes existence; a later algebraic equation may identify the limit only after existence is justified.

6.4 An infinite series is a limit of finite totals

Definition

For terms $a_1,a_2,\ldots$, define the $N$th partial sum

$$S_N=\sum_{n=1}^{N}a_n.$$

The infinite series $\sum_{n=1}^{\infty}a_n$ converges to $S$ when the sequence $(S_N)$ converges to $S$. Its sum is that limit. If the partial sums do not have a finite limit, the series diverges. The notation is a question about a limit, not permission to finish an infinite number of operations.

A necessary condition for convergence is $a_n\to0$. If $S_N\to S$, then for $n\ge2$,

$$a_n=S_n-S_{n-1}\longrightarrow S-S=0.$$

Common mistake

Thus terms that fail to approach zero prove divergence. Terms that approach zero do not prove convergence.

Worked example

Worked example 4: distinguish terms from totals. The sequence $a_n=1/n$ converges to zero. Nevertheless its series diverges. Group a partial sum ending at $2^m$ as

$$1+\frac12+\left(\frac13+\frac14\right) +\left(\frac15+\cdots+\frac18\right)+\cdots.$$

The block from $2^{j-1}+1$ through $2^j$ has $2^{j-1}$ terms, each at least $1/2^j$, so each block contributes at least $1/2$. Consequently $S_{2^m}\ge1+m/2$, which is unbounded. These are finite groupings of selected partial sums, not an unexplained rearrangement of an infinite series. This proves divergence of the harmonic series.

6.5 Geometric series, all cases, and remainders

Proposition

Consider $a+ar+ar^2+\cdots$, with partial sum $S_N=\sum_{k=0}^{N-1}ar^k$. If $a=0$, every term is zero and the series converges to zero for every real $r$. Suppose now that $a\ne0$.

For $r\ne1$, Chapter 5 gives $S_N=a(1-r^N)/(1-r)$. When $|r|<1$, $r^N\to0$, and therefore

$$\sum_{k=0}^{\infty}ar^k=\frac a{1-r}.$$

When $|r|\ge1$, the terms $ar^k$ do not approach zero, so the series diverges. This includes $r=1$ and $r=-1$. Writing $a/(1-r)$ at $r=2$ gives an ordinary number but does not make it the sum of the divergent series.

For $|r|<1$, the exact remainder after the first $N$ terms is

$$R_N=S-S_N=\frac{ar^N}{1-r},\qquad |R_N|=\frac{|a|\,|r|^N}{|1-r|}.$$

The simpler estimate $|R_N|\le |a||r|^N/(1-|r|)$ also holds. The exact expression retains the sign and accounts correctly for negative ratios.

Worked example

Worked example 5: alternating geometric accumulation. For $a=6$ and $r=-1/3$, the sum is $6/(1+1/3)=9/2$. After four terms,

$$S_4=6-2+\frac23-\frac29=\frac{40}{9}.$$

The remainder is $(9/2)-(40/9)=1/18$, agreeing with $6(-1/3)^4/(4/3)$. The fourth partial sum is below the limit, while the fifth is above it.

Worked example

Worked example 6: an echo tolerance. An idealized signal has successive contributions $3,3/2,3/4,\ldots$ in one fixed signal unit. Assuming the linear superposition model applies, its limiting total is six units. The error after $N$ contributions is $6/2^N$. To make it at most $0.01$, require $2^N\ge600$. Since $2^9=512<600$ and $2^{10}=1024\ge600$, the smallest permitted $N$ is ten. A real echo model may cease to be accurate before infinitely many terms; the sum is a statement within the stated idealization.

6.6 Telescoping, comparison and an alternating example

Chapter 5 gives

$$S_N=\sum_{n=1}^{N}\frac1{n(n+1)}=1-\frac1{N+1}.$$

Taking the limit proves that the series sums to one. Its remainder is $1/(N+1)$, not the first omitted term $1/[(N+1)(N+2)]$. Confusing a remaining total with one remaining contribution can greatly understate an error.

For nonnegative terms, partial sums are nondecreasing. Hence their series converges exactly when the partial sums are bounded above. This follows from monotone bounded convergence. In particular, if $0\le a_n\le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges: every partial sum of $a_n$ lies between zero and the finite sum of the $b_n$ series. A finite initial segment can be treated separately if the comparison starts later.

For example, when $n\ge2$,

$$0<\frac1{n^2}\le\frac1{n(n-1)}=\frac1{n-1}-\frac1n.$$

Summing the upper bound from two to $N$ gives $1-1/N<1$. After adding the first term, partial sums of $\sum 1/n^2$ are below two. The series therefore converges, although this comparison does not calculate its exact sum. We do not need a general calculus-based series test.

There is also a useful elementary alternating result. If $b_n\ge0$, $b_{n+1}\le b_n$ and $b_n\to0$, then

$$b_1-b_2+b_3-b_4+\cdots$$

converges. Its error after $N$ terms has magnitude at most $b_{N+1}$. To justify this, its even partial sums are nondecreasing, with increments $b_{2m+1}-b_{2m+2}\ge0$, while its odd partial sums are nonincreasing, with increments $-b_{2m+2}+b_{2m+3}\le0$. Every even partial sum is at most every odd partial sum: they form nested lower and upper bounds. Both subsequences converge by monotone bounded convergence, and their gap $b_{2m+1}$ tends to zero, so their limits coincide. The common limit lies between any two consecutive partial sums; their distance is the first omitted term, proving the error bound.

For $b_n=1/n$, this proves convergence of $1-1/2+1/3-\cdots$, even though the all-positive harmonic series diverges. The criterion requires a nonincreasing sequence tending to zero, not merely an alternating sign. No exact value for this sum is needed here. We preserve the stated order of terms: arbitrary rearrangement is not justified by these elementary arguments.

6.7 Graded exercises

Foundation

Checkpoint

Exercise 1

1. Write the first five terms of $a_n=2-3(1/2)^n$ for $n\ge0$. Find its limit.

Checkpoint

Exercise 2

2. For $a_n=4+2/n$, find an integer $N$ ensuring $|a_n-4|<0.005$ for all $n\ge N$.

Checkpoint

Exercise 3

3. Which of $1/n$, $(-1)^n$, $n/(n+1)$ and $n$ converge to finite real numbers? Give their limits when they do.

Checkpoint

Exercise 4

4. Write the first four partial sums of the series with $a_n=(-1)^{n-1}$, $n\ge1$. Does the series converge?

Core

Checkpoint

Exercise 5

5. Find $\lim_{n\to\infty}(7n+3)/(3n-2)$ and explain why the quotient rule applies after normalization.

Checkpoint

Exercise 6

6. For $v_0=1$ and $v_{n+1}=(v_n+8)/3$, derive and verify an explicit formula, then find its limit.

Checkpoint

Exercise 7

7. Decide convergence and find the sum when it exists: (a) $\sum_{k=0}^{\infty}5(2/3)^k$; (b) $\sum_{k=0}^{\infty}4(-1)^k$; (c) $\sum_{k=0}^{\infty}0\cdot2^k$.

Checkpoint

Exercise 8

8. Sum $\sum_{n=1}^{\infty}1/[(n+2)(n+3)]$ and find the exact remainder after $N$ terms.

Checkpoint

Exercise 9

9. For $\sum_{k=0}^{\infty}2(1/3)^k$, find the smallest $N\ge1$ for which the error after $N$ terms is at most $1/100$.

Checkpoint

Exercise 10

10. Explain why $a_n\to0$ is not enough to prove $\sum a_n$ converges, and give one convergent and one divergent series whose positive terms tend to zero.

Synthesis

Checkpoint

Exercise 11

11. Prove convergence of $\sum_{n=1}^{\infty}1/(n^2+1)$ without evaluating its sum. Give a finite upper bound for its total.

Checkpoint

Exercise 12

12. Does $\sum_{n=1}^{\infty}(-1)^{n-1}/(2n-1)$ converge by the alternating criterion? How many terms suffice for an error at most $0.001$ under its remainder bound?

Checkpoint

Exercise 13

13. A recurring input adds $4$ units immediately and then $4(0.8)^k$ units after the $k$th delay, $k\ge1$. Find the idealized infinite total and the total through delay $m$. Be explicit about the number of terms.

Checkpoint

Exercise 14

14. A student claims that $w_{n+1}=w_n^2$, $w_0=2$, converges to $L$ and therefore $L=L^2$, so its limit is zero or one. Identify the logical error and determine the actual behavior.

Checkpoint

Exercise 15

15. Express $0.272727\ldots$ as an infinite geometric series and as a reduced fraction. Justify the convergence used.

Checkpoint

Exercise 16

16. Suppose $a_n\to L$. Prove the sequence $a_{n+1}$ has the same limit. Use this fact to justify the necessary zero-term condition for a convergent series, keeping the roles of $a_n$ and $S_n$ distinct.

6.8 Full solutions

Solution · Exercise 1

1. The terms are $-1,1/2,5/4,13/8,29/16$. Since $(1/2)^n\to0$, their limit is two. The first index is zero, so the initial term is not $1/2$.

Solution · Exercise 2

2. Require $2/n<1/200$, or $n>400$. Thus $N=401$ works; $N=400$ fails the strict condition at its first index.

Solution · Exercise 3

3. The first converges to zero. The second has even terms one and odd terms minus one, so no finite limit exists. The third equals $1-1/(n+1)$ and tends to one. The fourth grows without bound, so does not converge to a finite real number.

Solution · Exercise 4

4. The partial sums are $1,0,1,0$. They keep oscillating and do not converge. Also the individual terms do not tend to zero, which independently rules out series convergence.

Solution · Exercise 5

5. Divide numerator and denominator by $n$ to obtain $(7+3/n)/(3-2/n)$. The limits of these expressions are seven and three, with a nonzero denominator limit. The quotient tends to $7/3$.

Solution · Exercise 6

6. The fixed point solves $v=(v+8)/3$, so it is four. Subtracting gives $v_{n+1}-4=(v_n-4)/3$. Therefore $v_n=4-3(1/3)^n$. At zero this gives one; substituting the formula yields the next term of the same form. Since $(1/3)^n\to0$, the limit is four. The verification, not the fixed-point equation alone, establishes the behavior.

Solution · Exercise 7

7. (a) The ratio has magnitude below one, so the sum is $5/(1-2/3)=15$. (b) Terms alternate between four and minus four and fail to tend to zero, so the series diverges. (c) Every term is zero, so the sum is zero; the zero-first-coefficient case does not require $|r|<1$.

Solution · Exercise 8

8. A term equals $1/(n+2)-1/(n+3)$. Thus $S_N=1/3-1/(N+3)$ and the infinite sum is $1/3$. The exact remaining total is $1/(N+3)$.

Solution · Exercise 9

9. The sum is three and the exact error is $3(1/3)^N=3^{1-N}$. At $N=5$ the error is $1/81>1/100$; at $N=6$ it is $1/243<1/100$. Since these errors decrease, the smallest $N$ is six.

Solution · Exercise 10

10. The harmonic series has terms $1/n\to0$ but its partial sums are unbounded by the block argument. The series $\sum_{n=1}^{\infty}1/[n(n+1)]$ also has positive terms tending to zero, but its partial sums tend to one. The necessary condition does not distinguish them.

Solution · Exercise 11

11. The first term is $1/2$. For $n\ge2$, $0<1/(n^2+1)\le1/n^2\le1/[n(n-1)]$. The comparison telescope after the first term has total one, so all partial sums are at most $3/2$. Nonnegative terms make them nondecreasing; bounded monotone convergence proves a finite sum, no greater than $3/2$.

Solution · Exercise 12

12. Here $b_n=1/(2n-1)$ is positive, decreases and tends to zero, so the series converges. The error after $N$ terms is at most $b_{N+1}=1/(2N+1)$. To make this at most $1/1000$, require $2N+1\ge1000$, hence integer $N\ge500$. This is a sufficient bound, not a proof that fewer terms can never achieve that actual error.

Solution · Exercise 13

13. The total is $4/(1-0.8)=20$ units. Through delay $m$ includes indices $k=0,\ldots,m$, so it has $m+1$ terms and total $20[1-(0.8)^{m+1}]$. The omitted total is $20(0.8)^{m+1}$.

Solution · Exercise 14

14. A fixed-point equation constrains a limit only if convergence has already been established and the passage to the limit is valid. Here $w_n=2^{2^n}$, as follows from the initial value and repeated squaring. In particular $w_n\ge2^{n+1}$ for $n\ge0$, so it grows without bound. There is no finite $L$ to insert into the equation.

Solution · Exercise 15

15. The decimal is $27/100+27/100^2+27/100^3+\cdots$. Its first term is $27/100$ and ratio $1/100$, so it converges to $(27/100)/(1-1/100)=27/99=3/11$. A finite truncation leaves an explicitly vanishing geometric tail.

Solution · Exercise 16

16. Given $\varepsilon>0$, choose $N$ with $|a_n-L|<\varepsilon$ for all $n\ge N$. Then for all $n\ge N$, also $n+1\ge N$, so $|a_{n+1}-L|<\varepsilon$. Now if the partial sums of a series satisfy $S_n\to S$, the shifted sequence $S_{n-1}$ also tends to $S$ on indices $n\ge2$: choose $n$ one larger than a tail threshold. Thus the terms $a_n=S_n-S_{n-1}$ tend to zero by the difference limit rule. It is convergence of the partial sums that supplies the conclusion about the terms, not the reverse.

Summary

A sequence converges when its entire tail lies within every prescribed tolerance of one finite number. An infinite series converges when the sequence of its finite partial sums does so. Geometric convergence requires $|r|<1$ unless every term is zero. Terms tending to zero are necessary but not sufficient for series convergence. Telescoping and comparison derive conclusions from finite partial sums; alternating monotone terms give another elementary convergence result with a first-omitted-term bound. An approximation is justified by its remaining total, not just by a small last included contribution.

Connections. Chapter 5 supplies the finite formulas. Chapter 7 justifies recurrence formulas and inequalities by induction. Chapter 9's logarithms can help solve a geometric error inequality when powers cannot be compared conveniently by integers.