What is

$\displaystyle \lim_{n\to\infty} \frac{1}{2^n}?$

You may have encountered sequences long ago in earlier courses and they seemed very difficult. You know much more mathematics now, so they will probably seem easier. On the other hand, we're going to go very quickly.

We will completely skip several topics from Chapter 11. I will try to make what we skip clear. Note that the homework has been modified to reflect the omitted topics.

A sequence is an ordered list of numbers. These numbers may be real, complex, etc., etc., but in this book we will focus entirely on sequences of real numbers. For example,

$\displaystyle \frac{1}{2},
\frac{1}{128}, \ldots, \frac{1}{2^n},\ldots

Since the sequence is ordered, we can view it as a function with domain the natural numbers $ =1,2,3,\ldots$.

Definition 6.1.1 (Sequence)   A sequence $ \{a_n\}$ is a function $ a:\N\to \mathbb{R}$ that takes a natural number $ n$ to $ a_n = a(n)$. The number $ a_n$ is the $ n$th term.

For example,

$\displaystyle a(n) = a_n = \frac{1}{2^n},$

which we write as $ \{\frac{1}{2^n}\}$. Here's another example:

$\displaystyle \left(b_n\right)_{n=1}^{\infty} = \left(\frac{n}{n+1}\right)_{n=1}^{\infty}
= \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots

Example 6.1.2   The Fibonacci sequence $ \left(F_n\right)_{n=1}^{\infty}$ is defined recursively as follows:

$\displaystyle F_1 = 1,   F_2 = 1,   F_n = F_{n-2} + F_{n-1}$   for $n &ge#geq;3$$\displaystyle .

Let's return to the sequence $ \left(\frac{1}{2^n}\right)_{n=1}^{\infty}$. We write $ \lim_{n\to\infty} \frac{1}{2^n} = 0$, since the terms get arbitrarily small.

Definition 6.1.3 (Limit of sequence)   If $ \left(a_n\right)_{n=1}^{\infty}$ is a sequence then that sequence converges to $ L$, written $ \lim_{n\to\infty} a_n = L$, if $ a_n$ gets arbitrarily close to $ L$ as $ n$ get sufficiently large. SECRET RIGOROUS DEFINITION: For every $ \varepsilon>0$ there exists $ B$ such that for $ n\geq B$ we have $ \vert a_n - L\vert<\varepsilon$.

This is exactly like what we did in the previous course when we considered limits of functions. If $ f(x)$ is a function, the meaning of $ \lim_{x\to\infty} f(x) = L$ is essentially the same. In fact, we have the following fact.

Proposition 6.1.4   If $ f$ is a function with $ \lim_{x\to\infty} f(x) = L$ and $ \left(a_n\right)_{n=1}^{\infty}$ is the sequence given by $ a_n = f(n)$, then $ \lim_{n\to\infty} a_n = L$.

As a corollary, note that this implies that all the facts about limits that you know from functions also apply to sequences!

Example 6.1.5  

$\displaystyle \lim_{n\to\infty} \frac{n}{n+1} = \lim_{x\to\infty} \frac{x}{x+1} = 1

Example 6.1.6   The converse of Proposition 6.1.4 is false in general, i.e., knowing the limit of the sequence converges doesn't imply that the limit of the function converges. We have $ \lim_{n\to\infty} \cos(2\pi n) = 1$, but $ \lim_{x\to\infty} \cos(2\pi x)$ diverges. The converse is OK if the limit involving the function converges.

Example 6.1.7   Compute $ \displaystyle \lim_{n\to\infty} \frac{n^3 + n + 5}{17n^3 - 2006n + 15}$. Answer: $ \frac{1}{17}$.

William Stein 2006-03-15