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.

To resolve the paradox, it must be true that \[\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \dfrac{1}{64} + \ldots = 1.\] 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 \(1+1+1+1+\ldots\), 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.

A sequence is simply a list of numbers. Here, we're only interested in *infinite* sequences: \[a_1, a_2, a_3, a_4, \ldots, a_n, \ldots\]

An infinite sequence is denoted by \[\{a_n\} \text{ or } \bigg\{a_n\bigg\}_{n=1}^\infty\] 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 1 (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 \(a_n\) without the braces refers to just the nth term of the sequence.

Write the first few terms of the sequence defined by \[a_n = 2n-1.\]

Remember, unless otherwise specified, \(n=1\) for the first term. Start with \(n=1\) and evaluate \(2n-1\) for each subsequent value of \(n\):

\[\ans{\{a_n\} = 1, 3, 5, 7, 9, \ldots}\]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 \[a_n = \dfrac{(-1)^n (n+1)}{3^n}.\]

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 \((-1)^n\).

Write the first few terms of the sequence defined by \[a_n = \cos \dfrac{n\pi}{6}.\]

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: \[\{a_n\} = \dfrac{5}{3}, \dfrac{10}{9}, \dfrac{15}{27}, \dfrac{20}{81}, \ldots\]

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

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

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 \(n\) in the formula.

\[\ans{a_n = \dfrac{5n}{3^n}}\]Find a formula for the nth term of the following sequence: \[\{a_n\} = \dfrac{1}{7}, -\dfrac{4}{11}, \dfrac{9}{15}, -\dfrac{16}{19}, \dfrac{25}{23}, \ldots\]

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

Next, look at the numerator: these are all perfect squares, so we'll have \(n^2\) in the numerator. Finally, the denominators are increasing by 4 each time, so there will be a \(4n\) in the expression. However, it should be \(4n+3\) in total, so that the first term will have 7.

\[\ans{a_n = \dfrac{(-1)^{n+1}\ n^2}{4n+3}}\]- Find a formula for the nth term of the following sequence: \[\{a_n\} = 1, \dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \dfrac{1}{9}, \ldots\]
- Find a formula for the nth term of the following sequence: \[\{a_n\} = 1, -\dfrac{2}{3}, \dfrac{4}{9}, -\dfrac{8}{27}, \ldots\]

\(a_n = \dfrac{1}{2n-1}\)

\(a_n = \dfrac{(-1)^{n+1}\ 2^{n-1}}{3^{n-1}}\)

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 \[\{a_n\} = 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, \ldots\]

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

As \(n \to \infty\), \(\dfrac{1}{n} \to 0\). 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).

A sequence \(\{a_n\}\) has limit \(L\), written \[\lim_{n \to \infty} a_n = L \hspace{0.5in} \text{ or } \hspace{0.5in} a_n \to L \text{ as } n \to \infty,\] if we can make \(a_n\) as close to \(L\) as we want by making \(n\) sufficiently large.

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

A sequence \(\{a_n\}\) has limit \(L\) if for every \(\epsilon > 0\) there exists an integer \(N\) such that if \(n > N\) then \(|a_n - L| < \epsilon\).

If \(\displaystyle\lim_{x \to \infty} f(x) = L\) and \(f(n)=a_n\) when \(n\) is a natural number, then \(\displaystyle\lim_{n \to \infty} a_n = L\).

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: \[\{a_n\} = \left\{\dfrac{\ln n}{n}\right\}\]

Start by transitioning to the limit of the corresponding function (this is necessary because we'll need to use L'Hopital's Rule):

\[\lim_{n \to \infty} \dfrac{\ln n}{n} = \lim_{x \to \infty} \dfrac{\ln x}{x}\]Since this is an indeterminate form, we can use L'Hopital's Rule:

\[\lim_{x \to \infty} \dfrac{\ln x}{x} = \lim_{x \to \infty} \dfrac{1/x}{1} = 0\]Therefore, \[\ans{\lim_{n \to \infty} \left\{\dfrac{\ln n}{n}\right\} = 0,}\] so this sequence converges.

- Determine whether the following sequence converges, and if so, the limit of the sequence. \[1-\left(\dfrac{1}{4}\right)^n\]
- Determine whether the following sequence converges, and if so, the limit of the sequence. \[\dfrac{n^3}{n^3+1}\]
- Determine whether the following sequence converges, and if so, the limit of the sequence. \[\dfrac{n^3}{n+1}\]
- Determine whether the following sequence converges, and if so, the limit of the sequence. \[\dfrac{3+5n^2}{n+n^2}\]
- Determine whether the following sequence converges, and if so, the limit of the sequence. \[0,1,0,0,1,0,0,0,1,\ldots\]

Converges to \(1\)

Converges to \(1\)

Diverges

Converges to \(5\)

Diverges

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

If \(\displaystyle\lim_{n \to \infty} |a_n| = 0\), then \(\displaystyle\lim_{n \to \infty} a_n = 0\).

Note: this is only true if the limit of the absolute values is 0; if it's anything else, this theorem doesn't hold. For example, consider the sequence \(\{a_n\} = \{(-1)^n\}\). The limit of the absolute values of this sequence is 1, and the sequence diverges, alternating between 1 and -1 forever.

Evaluate \(\displaystyle\lim_{n \to \infty} \dfrac{(-1)^n}{n^2}\).

Based on the absolute value theorem, this sequence converges to 0:

\[\lim_{n \to \infty} \left|\dfrac{(-1)^n}{n^2}\right| = \lim_{n \to \infty} \dfrac{1}{n^2} = 0 \longrightarrow \ans{\lim_{n \to \infty} \dfrac{(-1)^n}{n^2} = 0}\]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 \(a_n \leq b_n \leq c_n\) for \(n\) greater than some integer \(n_0\) and \(\displaystyle\lim_{n \to \infty} a_n = \displaystyle\lim_{n \to \infty} c_n = L\), then \(\displaystyle\lim_{n \to \infty} b_n = L\).

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

Evaluate \(\displaystyle\lim_{n \to \infty} \dfrac{1}{n!}\).

To use the squeeze theorem, we need a lower sequence and an upper sequence. First, notice that \(1/n!\) is always greater than 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 \(n! > 2^n\) for \(n>3\), then we can see that \(1/n! < 1/2^n\) for \(n>3\). Therefore, since \[0 < \dfrac{1}{n!} < \dfrac{1}{2^n} \text{ for } n>3\] and \[\lim_{n \to \infty} \{0\} = \lim_{n \to \infty} \left\{\dfrac{1}{2^n}\right\} = 0 \longrightarrow \ans{\lim_{n \to \infty} \left\{\dfrac{1}{n!}\right\} = 0}\]

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

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

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 \[\dfrac{3}{(n+1)+5} \leq \dfrac{3}{n+5}\] (simply cross-multiply and simplify the inequality until it is obvious). More intuitively, the numerator is constant, and as \(n\) 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, cannnot become negative.

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