Infinite Sequences

Here's an ancient paradox, known as one of Zeno's Paradoxes: if Achilles wants to go a mile, he must first travel half the distance there. So far so good, but now to get from the halfway point to the end, he must travel half of that distance, or a quarter of the full distance. Then he must go an eighth of the full distance, a sixteenth, and so on; there will always be half of the remaining distance that he must travel. The paradox is this: if Achilles has to travel an infinite number of small distances, how can he ever reach his destination? Of course, motion is possible, so there must be an answer.

CC-BY-SA Martin Grandjean

To resolve the paradox, it must be true that 12+14+18+116+132+164+…=1.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\ldots=1.

This is an example of an infinite series, the topic of this last unit of Calculus II. The mind-bending part is that we can add up an infinite list of numbers and get a finite result; this series is therefore what we call a convergent series. Obviously, not every infinite series converges; if we added 1+1+1+1+…,1+1+1+1+\ldots, that would tend toward infinity, or diverge.

Before we get to infinite series, though, we'll start with a short section on infinite sequences; it's important to keep sequences and series distinct in your mind; many students, when they first see this, confuse the two. In short, a sequence is a list of numbers, and a series is a list of numbers that are added together. Just remember, what we learn here about sequences will not hold for series; there are a few cases where it is easy to confuse the two.

Definition and Notation

A sequence is simply a list of numbers. Here, we're only interested in infinite sequences: a1,a2,a3,a4,…,an,…a_1,a_2,a_3,a_4,\ldots,a_n,\ldots

Notation

An infinite sequence is denoted by {an} or {an}n∞=1{a_n} \textrm{ or } \bigg{a_n\bigg}_n^{\infty}=1

The second one is more explicit about the sequence being infinite, but since we're only going to deal with infinite sequences, and the index on ours will always start at 11 (unless otherwise specified), I'll usually just use the notation on the left. The curly braces indicate that we're talking about the whole sequence; just writing ana_n without the braces refers to just the nnth term of the sequence.

Write the first few terms of the sequence defined by an=2n−1.a_n=2n−1.

Solution

Remember, unless otherwise specified, n=1n=1 for the first term. Start with n=1n=1 and evaluate 2n−12n−1 for each subsequent value of nn: an=1,3,5,7,9,…\ans{{a_n}=1,3,5,7,9,\ldots}

Note that we just listed enough terms to make the pattern in the sequence clear; this is the list of the odd numbers.

Write the first few terms of the sequence defined by an=(−1)n(n+1)3n.a_n=\dfrac{(−1)^n(n+1)}{3^n}.

Solution

an=−23,39,−427,581,…\ans{{a_n}=−\dfrac{2}{3},\dfrac{3}{9},−\dfrac{4}{27},\dfrac{5}{81},\ldots}

This is an example of an alternating sequence, which is one where the terms flip back and forth from positive to negative; this, of course, is due to the (−1)n(−1)^n.

Write the first few terms of the sequence defined by an=cos⁡nπ6a_n=\cos \dfrac{n\pi}{6}

Solution

an=32,12,0,−12,−32,−1,…\ans{{a_n}=\dfrac{\sqrt{3}}{2},\dfrac{1}{2},0,−\dfrac{1}{2},−\dfrac{\sqrt{3}}{2},−1,\ldots}

Finding the Formula for a Sequence

Sometimes it's necessary to look at a sequence written out as a list, and compress that into a formula.

Find a formula for the nth term of the following sequence: {an}=53,109,1527,2081,…{a_n}=\dfrac{5}{3},\dfrac{10}{9},\dfrac{15}{27},\dfrac{20}{81},\ldots

Solution

Look at the denominator first: each time nn increases by one (going to the next term), the denominator gets multiplied by 33; this repeated multiplication leads to an exponent. The denominator can be described, therefore, by 3n.3^n.

What about the numerator? Each subsequent term has 55 added to it; this repeated addition leads to multiplication. You should verify that it makes sense that the numerator can be described by 5n.5n.

In general, if each term gets multiplied by something to get to the next term, there will be an exponent in the formula, and if each term has something added to it to get the next term, there will be a multiple of nn in the formula. an=5n3n\ans{a_n=\dfrac{5n}{3^n}}

Find a formula for the nth term of the following sequence: {an}=17,−411,915,−1619,2523,…{a_n}=\dfrac{1}{7},−\dfrac{4}{11},\dfrac{9}{15},−\dfrac{16}{19},\dfrac{25}{23},\ldots

Solution

First of all, notice that this is an alternating sequence, but we won't use (−1)n,(−1)^n, because that would make the first term negative. We want to shift the negatives to the even terms; to do this, we'll use (−1)n+1.(−1)^{n+1}.

Next, look at the numerator: these are all perfect squares, so we'll have n2n^2 in the numerator. Finally, the denominators are increasing by 44 each time, so there will be a 4n4n in the expression. However, it should be 4n+34n+3 in total, so that the first term will have 7.7. an=(−1)n+1n24n+3\ans{a_n=(−1)^{n+1} \dfrac{n^2}{4n+3}}

  1. Find a formula for the nth term of the following sequence: {an}=1,13,15,17,19,…{a_n}=1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9},\ldots

    2. an=12n−1a_n=\dfrac{1}{2n−1}

  2. Find a formula for the nth term of the following sequence: {an}=1,−23,49,−827,…{a_n}=1,−\dfrac{2}{3},\dfrac{4}{9},−\dfrac{8}{27},\ldots

    3. an=(−1)n+1 2n−13n−1a_n=\dfrac{(−1)^{n+1}\ 2^{n−1}}{3^{n−1}}

Of course, there are some sequences that have a clear pattern, but it's not easy to find a formula for it. An example is a sequence like an=1,0,1,0,0,1,0,0,0,1,…{a_n}=1,0,1,0,0,1,0,0,0,1,\ldots

The Limit of an Infinite Sequence

The limit of a sequence is fairly intuitive. Take, for example, the sequence defined by an=1n.a_n=\dfrac{1}{n}. If we plot this sequence, we can see the trend.

Limit of 1/n

As n→∞,n \to \infty, 1n→0.\dfrac{1}{n} \to 0. This, of course, is similar to the limit of a function (this is not accidental, since a sequence can be thought of as a function from the natural numbers to the real numbers).

Definition of Limit

A sequence an{a_n} has limit L,L, written lim⁡n→∞an=L or an→L as n→∞,\lim_{n \to \infty} a_n=L \textrm{ or } a_n \to L \textrm{ as } n \to \infty, if we can make ana_n as close to LL as we want by making nn sufficiently large.

If this limit exists, the sequence is convergent. Otherwise, it is divergent.

More Mathy Definition

A sequence an{a_n} has limit LL if for every ϵ>0\epsilon > 0 there exists an integer NN such that if n>Nn>N then ∣an−L∣<ϵ|a_n−L| < \epsilon.

Theorem

If lim⁡x→∞f(x)=L\displaystyle\lim_{x \to \infty} f(x)=L and f(n)=anf(n)=a_n when nn is a natural number, then lim⁡n→∞an=L.\displaystyle\lim_{n \to \infty} a_n=L.

This theorem basically states that we can use what we know about limits with functions to evaluate limits of sequences.

Find the limit of the following sequence: {an}={ln⁡nn}{a_n}=\left{\dfrac{\ln n}{n}\right}

Solution

Start by transitioning to the limit of the corresponding function (this is necessary because we'll need to use L'Hopital's Rule): lim⁡n→∞ln⁡nn=lim⁡x→∞ln⁡xx\lim_{n \to \infty} \dfrac{\ln n}{n} = \lim_{x \to \infty} \dfrac{\ln x}{x}

Since this is an indeterminate form, we can use L'Hopital's Rule: lim⁡x→∞ln⁡xx=lim⁡x→∞1/x1=0\lim_{x \to \infty} \dfrac{\ln x}{x} = \lim_{x \to \infty} \dfrac{1/x}{1}=0

Therefore, lim⁡n→∞{ln⁡nn}=0\ans{\lim_{n \to \infty}\left{\dfrac{\ln n}{n}\right}=0} so this sequence converges.

  1. Determine whether the following sequence converges, and if so, find the limit of the sequence. 1−(14)n1−\left(\dfrac{1}{4}\right)^n

    2. Converges to 11

  2. Determine whether the following sequence converges, and if so, find the limit of the sequence. n3n3+1\dfrac{n^3}{n^3+1}

    3. Converges to 11

  3. Determine whether the following sequence converges, and if so, find the limit of the sequence. n3n+1\dfrac{n^3}{n+1}

    4. Diverges

  4. Determine whether the following sequence converges, and if so, find the limit of the sequence. 3+5n2n+n2\dfrac{3+5n^2}{n+n^2}

    5. Converges to 55

  5. Determine whether the following sequence converges, and if so, find the limit of the sequence. 0,1,0,0,1,0,0,0,1,…0,1,0,0,1,0,0,0,1,\ldots

    6. Diverges

We'll close this section with three theorems about the convergence of sequences.

Theorem 1: Absolute Value Theorem

If lim⁡n→∞∣an∣=0,\displaystyle\lim_{n \to \infty} |a_n|=0, then lim⁡n→∞an=0.\displaystyle\lim_{n \to \infty} a_n=0.

Note: this is only true if the limit of the absolute values is 00; if it's anything else, this theorem doesn't hold. For example, consider the sequence {an}={(−1)n}.{a_n}={(−1)^n}. The limit of the absolute values of this sequence is 1,1, and the sequence diverges, alternating between 11 and −1-1 forever.

Evaluate lim⁡n→∞(−1)nn2.\displaystyle\lim_{n \to \infty}\dfrac{(−1)^n}{n^2}.

Solution

Based on the absolute value theorem, this sequence converges to 0: lim⁡n→∞∣(−1)nn2∣=lim⁡n→∞1n2=0⟶lim⁡n→∞(−1)nn2=0\lim_{n \to \infty} \bigg|\dfrac{(−1)^n}{n^2}\bigg| = \lim_{n \to \infty} \dfrac{1}{n^2}=0 \longrightarrow \ans{\lim_{n \to \infty} \dfrac{(−1)^n}{n^2}=0}

Theorem 2: Squeeze Theorem

Pretend you're driving on a highway, flanked on either side by 18-wheelers. If both of those trucks decide to take the next exit, guess what: you're taking the exit with them, whether you like it or not. That's essentially what the Squeeze Theorem says.

If an≤bn≤cna_n \leq b_n \leq c_n for nn greater than some integer n0n_0 and lim⁡n→∞an=lim⁡n→∞cn=L\displaystyle\lim_{n \to \infty} a_n = \displaystyle\lim_{n \to \infty} c_n=L, then lim⁡n→∞bn=L.\displaystyle\lim_{n \to \infty} b_n=L.

Squeeze Theorem

The hard part is usually figuring out what sequences to use as the upper and lower bounding sequences.

Evaluate lim⁡n→∞1n!.\displaystyle\lim_{n \to \infty} \dfrac{1}{n!}.

Solution

To use the squeeze theorem, we need a lower sequence and an upper sequence. First, notice that 1/n!1/n! is always greater than 0,0, so we can use the sequence of zeros as the lower bound.

The upper bound is a bit trickier, but if we note that n!>2nn!>2n for n>3,n>3, then we can see that 1/n!<1/2n1/n!<1/2n for n>3.n>3. Therefore, since 0<1n!<12n for n>30 < \dfrac{1}{n!} < \dfrac{1}{2n} \textrm{ for } n>3 and lim⁡n→∞{0}=lim⁡n→∞{12n}=0\lim_{n \to \infty}{0} = \lim_{n \to \infty}{\dfrac{1}{2n}}=0 we can use the Squeeze Theorem to conclude that lim⁡n→∞{1n!}=0\ans{\lim_{n \to \infty}\left{\dfrac{1}{n!}\right}=0}

Theorem 3: Monotone Sequence Theorem

Definition of Monotone

A sequence is increasing if an+1≥ana_{n+1} \geq a_n for every n≥1.n \geq 1. A sequence is decreasing if an+1≤ana_{n+1} \leq a_n for every n≥1.n \geq 1. In either case, the sequence is monotonic. Basically, monotonic just means "going consistently in one direction."

Theorem

Every monotone bounded sequence is convergent.

More precisely, every increasing sequence that is bounded above is convergent, and every decreasing sequence that is bounded below is convergent.

Prove that {3n+5}\left{\dfrac{3}{n+5}\right} is convergent using the monotone sequence theorem.

Solution

We have to prove two things: that the sequence is monotonic (this one happens to be decreasing) and bounded (below in this case).

We can rigorously show that the sequence is decreasing by proving that 3(n+1)+5≤3n+5\dfrac{3}{(n+1)+5} \leq \dfrac{3}{n+5} (simply cross-multiply and simplify the inequality until it is obvious). More intuitively, the numerator is constant, and as nn increases, so does the denominator, so it should be clear that this sequence is decreasing.

This sequence is bounded below by 0, since the terms of the sequence, though they get smaller, cannot become negative.

Therefore, by the monotone sequence theorem, this sequence is convergent.