Sequences

A sequence is an ordered list of numbers. We are concerned with sequences that have a formula for each term in the list. The notation we will use for sequences is $\{a_n\}^{n=\infty}_{n=k}$ where $a_n=f(n)$. Fot a specific example, consider the sequence

$$\{a_n\displaystyle \}^{n=\infty}_{n=0}\;\;$$where$$\;\; a_n=(-1)^{n+1}\displaystyle \frac{n^2}{2n-1}$$

Then we get

$$a_0 = (-1)^{1}\displaystyle \frac{0}{-1}=0$$

$$a_1 = (-1)^{1+1}\displaystyle \frac{1^2}{2-1}=1$$

$$a_2 = (-1)^{2+1}\displaystyle \frac{2^2}{4-1}=-\displaystyle \frac{4}{3}$$

$$a_3 = (-1)^{3+1}\displaystyle \frac{3^2}{6-1}=\displaystyle \frac{9}{5}$$

$$\vdots$$

$$a_n = (-1)^{n+1}\displaystyle \frac{n^2}{2n-1}$$

$$\vdots$$

We ae interested in whether a sequence of numbers approaches a single value as $n$ gets very large. In other words, does $\lim_{n\to\infty}a_n$ exist? To evaluate such a limit, realize that each sequence is associated with a continuous function. So the points of the sequence $\{a_n\}$ where $a_n=n^2+1$ are points on the smooth curve $f(x)=x^2+1$.

To evaluate the limit of a sequence we simply find the limit of the function associated with this sequence. The two limits will be the same. One problem arises when the sequence has alternating negative and positive terms such as in the first example. Then to evaluate this limit we must associate the positive terms with one function and the negative terms with that functions negative. For a simlar example consider,

$$a_n=(-1)^{n+1}\displaystyle \frac{6n}{2n-1}=\left\{ \begin{ar... ...] -\displaystyle \frac{6n}{2n-1}&n-even\\ \end{array}\right .$$

So for even terms the points lie on the curve $f(x)=-6x/(2x-1)$ and for odd $n$ the terms lie on the function $g(x)=6x/(2x-1)$. Now we must evaluate both limits.

$$\displaystyle \lim_{x\to\infty}-\displaystyle \frac{6x}{2x-1}=-3$$

$$\displaystyle \lim_{x\to\infty}\displaystyle \frac{6x}{2x-1}=3$$

The positive terms are approaching 3 and the negative terms are approaching -3 and so the limit of the sequence does not exist!

$$\displaystyle \lim_{n\to\infty}a_n\;\mathbf{d.n.e.}$$