Part II · Limits, Continuity, and the Calculus Foundation

Chapter 5 — Sequences and Infinite Processes

Why This Matters

Many engineering ideas are iterative: a sampled signal is a sequence, a numerical algorithm produces successive approximations, and a transient circuit approaches a steady state. Limits give a precise language for “approaches.”

5.1 Sequences

A sequence is a function whose inputs are positive integers. We write

(a_n)_{n=1}^{\infty}=a_1,a_2,a_3,\ldots

For example, $$a_n=1/n$$ gives $1,1/2,1/3,\ldots$ . Unlike a continuous graph, visualise a sequence as isolated dots above the integer positions $1,2,3,\ldots$ .

A sequence may be explicit, such as $$a_n=3n-1$$ , or recursive, such as

a_{n+1}=\frac12\left(a_n+\frac2{a_n}\right),\qquad a_1>0.

The recursive rule above is Newton’s iteration for $\sqrt2$ .

A sequence is discrete data. Its input is the positive integer index, so a plotted sequence is a collection of dots rather than a continuous curve. The notation a_n names one term; it does not mean that every real horizontal position is an allowed input.

(a_n)_{n=1}^{\infty}=a_1,a_2,\ldots,\qquad a_n=\frac1n

Discrete terms of a sub n equals one over n, plotted only at integer indices.

5.2 Limits of sequences

We say $$a_n$$ converges to $$L$$ , written $a_n\to L$ , if its terms can be made as close to $$L$$ as desired by going sufficiently far along the sequence. Formally, for every $\varepsilon>0$ there is an integer $$N$$ such that

n\ge N\implies |a_n-L|<\varepsilon.

The value $$L$$ need not appear as a sequence term. For $$a_n=1/n$$ , no term is zero, but terms eventually fit inside every band $(-\varepsilon,\varepsilon)$ around zero.

Example — An $\varepsilon$ – $$N$$ proof

Prove $1/n\to0$ . Given $\varepsilon>0$ , choose an integer $N>1/\varepsilon$ . For $n\ge N$ ,

\left|\frac1n-0\right|=\frac1n\le\frac1N<\varepsilon.

This proves convergence. The proof works because it tells us exactly how far we must go for any requested accuracy.

The index depends on the requested accuracy. For 1/n, choose an integer beyond 1/ε. Then every later denominator is still larger, so every later term lies in the same ε-band; this proves an eventual statement, not merely a pattern among early dots.

N>\frac1{\varepsilon},\ n\ge N\Longrightarrow\left|\frac1n-0\right|\le\frac1N<\varepsilon

State a positive tolerance ε.
Choose an integer N greater than 1/ε.
Use n ≥ N to bound 1/n by 1/N, then by ε.

For ε equals 0.05, the terms one over n enter the strict band after n equals 20; n equals 21 is the first integer inside it.

5.3 Boundedness and monotonicity

A sequence is bounded above by $$M$$ if $a_n\le M$ for all $$n$$ , and bounded below by $$m$$ if $a_n\ge m$ for all $$n$$ . It is non-decreasing if $a_{n+1}\ge a_n$ and non-increasing if $a_{n+1}\le a_n$ .

The theorem guarantees existence; it does not by itself find the value of the limit. For a recursive sequence, first prove the appropriate bounds and monotonicity, then take limits on both sides of the recursion.

Bounded monotonic motion has somewhere to settle. A non-decreasing real sequence with an upper bound converges, and the reversed statement holds for a non-increasing sequence with a lower bound. This theorem guarantees existence only; a separate limit equation or estimate is needed to identify the value.

a_{n+1}\ge a_n,\ a_n\le M\Longrightarrow(a_n)\text{ converges}

Prove the proposed bound holds for every term.
Compare consecutive terms to establish monotonicity.
Use the theorem for existence before solving a fixed-point equation for the limit.

5.4 Series and partial sums

Given a sequence $$(a_n)$$ , the expression

\sum_{n=1}^{\infty}a_n

is a series. Its meaning is the limit of partial sums

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

Do not confuse $a_n\to0$ with convergence of the series. It is necessary for a series to converge, but it is not sufficient: the harmonic series $\sum_{n=1}^{\infty}1/n$ diverges even though $1/n\to0$ .

A series is defined through a new sequence. The individual terms a_n and the partial sums S_N answer different questions. Term convergence to zero is necessary, but only convergence of the partial-sum sequence makes the infinite addition finite.

S_N=\sum_{n=1}^{N}a_n,\qquad a_n\to0\not\Longrightarrow S_N\text{ converges}

Partial sums separate geometric levelling from harmonic growth, even though both displayed term sequences shrink.

5.5 Sequence limit laws

If $a_n\to A$ and $b_n\to B$ , then sums, products, and scalar multiples behave as expected:

a_n+b_n\to A+B,\qquad a_nb_n\to AB, \qquad \frac{a_n}{b_n}\to\frac AB\quad(B\ne0).

For rational expressions in $$n$$ , divide numerator and denominator by the highest power of $$n$$ that occurs.

Some sequence limits are not rational expressions. A standard one is

\lim_{n\to\infty}a^{1/n}=1\qquad(a>0).

It says that repeatedly taking a root pulls every fixed positive number toward $$1$$ . One way to see the structure is to write $a^{1/n}=e^{(\ln a)/n}$ : the exponent tends to $$0$$ , and continuity of the exponential gives the result. This is not a substitution of infinity into an exponent; it is a composition of two ordinary limits.

Example — Dominant powers

\lim_{n\to\infty}\frac{5n^2-3n+1}{2n^2+7} =\lim_{n\to\infty}\frac{5-3/n+1/n^2}{2+7/n^2}=\frac52.

We did not substitute “ $\infty$ ” for $$n$$ . Infinity is not an ordinary real number; the division exposes terms whose limits are known.

Limit laws apply to known component limits. Dividing a rational expression by its highest power exposes constants and reciprocal powers whose limits are controlled. It is not substitution of infinity, and a quotient law still requires a non-zero limiting denominator.

\frac{5n^2-3n+1}{2n^2+7}=\frac{5-3/n+1/n^2}{2+7/n^2}\longrightarrow\frac52

Identify the highest power occurring in numerator and denominator.
Divide every term by that power.
Apply the sum and quotient laws only after checking the limiting denominator is non-zero.

Common Mistakes

Treating a sequence as a continuous curve rather than a discrete list.
Assuming a limit must equal some term.
Calling $$0/0$$ an answer instead of recognising an indeterminate form.
Applying a quotient law when the limiting denominator is zero.

5.6 Tails, oscillation, and partial sums

Convergence concerns the tail of a sequence: changing finitely many early terms does not change its limit. Picture a horizontal $\varepsilon$ -band around $$L$$ . Once every later dot lies in that band, the definition is met. Oscillation alone does not prevent convergence; what matters is whether the distance to one value shrinks.

Example — an oscillating sequence that converges

For $$a_n=(-1)^n/n$$ ,

|a_n-0|=\frac1n.

Given $\varepsilon>0$ , choose $N>1/\varepsilon$ . Every $n\ge N$ satisfies $|a_n|<\varepsilon$ , independent of its sign. Thus $a_n\to0$ .

Example — a fixed point needs an existence argument

Let $a_{n+1}=(a_n+4)/2$ with $$a_1=0$$ . A possible limit satisfies $$L=(L+4)/2$$ , giving $$L=4$$ . To justify taking that limit, observe $a_n\le4$ and

a_{n+1}-a_n=\frac{4-a_n}{2}\ge0.

The sequence is increasing and bounded above, so it converges; only then does the fixed-point calculation determine its limit.

Example — terms tend to zero but partial sums grow

Group the harmonic series:

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

Every group after the first is at least $$1/2$$ . The partial sums consequently exceed $$1+m/2$$ after $$m$$ such groups and are unbounded. Thus $1/n\to0$ does not imply $\sum1/n$ converges.

[Figure: Dots for

\(1/n\)

and

\((-1)^n/n\)

entering an

\varepsilon

-band, with partial-sum steps for a convergent geometric series contrasted with harmonic sums.]

Oscillation is harmless when its size shrinks. The signs of (−1)^n/n alternate, but its distance from zero is exactly 1/n, so its tail enters every epsilon band. In contrast, harmonic partial sums grow because grouped positive terms keep adding a fixed minimum amount.

\left|\frac{(-1)^n}{n}\right|=\frac1n\to0,\qquad H_N=\sum_{n=1}^{N}\frac1n\to\infty

Oscillating terms shrink toward zero while harmonic partial sums still rise; the displayed objects are not interchangeable.

5.7 A sequence is not the series built from it

The sequence $$a_n$$ is a list of individual contributions. The series $\sum a_n$ is a new object whose values are the partial sums $S_N=a_1+\cdots+a_N$ . A convergent sequence can therefore produce either a convergent or divergent series. The term test only provides a one-way decision: if $$a_n$$ fails to approach zero, the series diverges; if it does approach zero, another test is required.

Example — geometric partial sums approach a finite total

For $a_n=(1/2)^{n-1}$ , the terms approach $$0$$ . More importantly,

S_N=1+\frac12+\cdots+\left(\frac12\right)^{N-1} =2\left(1-\left(\frac12\right)^N\right).

Since $(1/2)^N\to0$ , $S_N\to2$ . The series converges because the partial sums have a limit, not merely because individual contributions shrink.

Example — a necessary condition that fails immediately

For $\sum_{n=1}^{\infty}(n+1)/(n+2)$ ,

\lim_{n\to\infty}\frac{n+1}{n+2}=1\ne0.

The terms do not even become small, so adding infinitely many of them cannot settle to a finite value. No ratio, comparison, or more elaborate test is needed.

Example — recursive convergence and an error estimate

If $a_{n+1}=(a_n+4)/2$ has limit $$4$$ , subtract $$4$$ from both sides:

a_{n+1}-4=\frac12(a_n-4).

Hence $|a_{n+1}-4|=\tfrac12|a_n-4|$ . Every iteration halves the current error. This connects convergence to a practical numerical stopping rule: after $$k$$ further steps, the error is at most $2^{-k}$ times the current error.

[Figure: One panel with sequence dots and another with their partial-sum points; show geometric partial sums leveling off while harmonic partial sums continue upward. A third panel shows iterative errors halving toward zero.]

Derive the infinite result from a finite identity. For a geometric sum, subtracting half the sum cancels every middle term. Only after obtaining the formula for S_N may we take its limit; for ratio one half, the third partial sum is 1.75 and the limiting total is two.

S_N=1+\frac12+\cdots+\left(\frac12\right)^{N-1}=2\left(1-\left(\frac12\right)^N\right)\to2

Write the finite sum S sub N.
Subtract one half S sub N so the middle terms cancel.
Take the limit only after the finite identity is established.

The finite geometric partial sums 1, 1.5, and S₃ equals 1.75 approach the total 2.

As more terms are added, the geometric partial sums 1, 1.5, 1.75, ... settle toward the limiting total 2.

Exercises

Decide whether $$a_n=(-1)^n$$ converges. Explain geometrically.
Prove that $3/(2n+1)\to0$ using the definition.
Find $\lim_{n\to\infty}(7n^3-n)/(2n^3+5n)$ .
Find the first five terms of $a_{n+1}=\frac12(a_n+3/a_n)$ with $$a_1=2$$ .
Explain why $a_n\to0$ is necessary but not sufficient for $\sum a_n$ to converge.
Prove $(-1)^n/n\to0$ using an $\varepsilon$ – $$N$$ argument.
Find a suitable $$N$$ so $$(3n-1)/(n+2)$$ is within $$0.01$$ of its limit.
Establish convergence of $a_{n+1}=(a_n+4)/2$ , $$a_1=0$$ , before finding its limit.
For $a_n=(1/2)^{n-1}$ , write $$S_N$$ and show $\lim S_N=2$ .
Use the term test to decide whether $\sum(n+1)/(n+2)$ can converge.

Why This Matters

5.1 Sequences

5.2 Limits of sequences

Example — An \(\varepsilon\)–\(N\) proof

5.3 Boundedness and monotonicity

5.4 Series and partial sums

5.5 Sequence limit laws

Example — Dominant powers

Common Mistakes

5.6 Tails, oscillation, and partial sums

Example — an oscillating sequence that converges

Example — a fixed point needs an existence argument

Example — terms tend to zero but partial sums grow

5.7 A sequence is not the series built from it

Example — geometric partial sums approach a finite total

Example — a necessary condition that fails immediately

Example — recursive convergence and an error estimate

Exercises

Example — An $\varepsilon$ – $\(N\)$ proof