Limits

We now discuss the "long-term behaviour" of sequences. Consider the sequence

(a_n)=(a_1, a_2, ...)

We assume that we have a recursive or explicit method to determine each term $a_n$ of the sequence. We are interested in knowing what happens to $a_n$ if we make $n$ larger and larger (hence, when we move to the right in the sequence). We have already discussed some examples at the beginning of the section. Let's define it in more detail.

Definition 1: Convergent sequence

We say that the sequence $(a_n)$ is convergent if the values $a_n$ get closer and closer (but not necessary touch) a single real number as we make $n$ larger and larger. This single number is often denoted by $a_\infty$ , and is called the limit of the sequence. This is then written as follows:

\lim_{n\rightarrow \infty} a_n = a_\infty

or as

a_n \xrightarrow[]{n\rightarrow \infty} a_\infty

(say: "the limit of the sequence $(a_n)$ is $a_\infty$ as $n$ approaches infinity", or "the sequence $(a_n)$ converges to $a_\infty$ if $n$ approaches infinity").

Definition 2: Divergent sequence

If the sequence $(a_n)$ does not converge, that is, if it does not approach a certain value, we call it divergent. There are two possible ways a sequence can be divergent.

Either the sequence does not approach a single value $a_\infty$ (e.g. by alternating forever between two values). Thus no (single) limit exists: $\lim_{n\rightarrow \infty} a_n \text{ does not exist}$
or the sequence increases without upper bound, in which case we write $\lim_{n\rightarrow \infty} a_n = \infty$ or $a_n \xrightarrow[]{n\rightarrow \infty} \infty$ (that is, the limit is $a_\infty = \infty$ )
or the sequence decreases without lower bound: $\lim_{n\rightarrow \infty} a_n = -\infty$ or $a_n \xrightarrow[]{n\rightarrow \infty} -\infty$ (that is, the limit is $a_\infty = -\infty$ ).

Example 1

Again the examples from the beginning of the chapter. Determine whether the sequences are convergent or divergent. If convergent, determine the limit of the sequence.

$(a_n)=(1,3,5,...)$
$(b_n)=(1, 0.5, 0.25, 0.125, ...)$
$(c_n)=(1,-1,1,-1,...)$
$(d_n)=(1,2,4,8,...)$

Also, give an example of a sequence that converges to $1.7$ .

Solution

Divergent, with $\lim_{n\rightarrow \infty} a_n = \infty$ .
Convergent, with $\lim_{n\rightarrow \infty} a_n = 0$ .
Divergent, the sequence does not approach a single number.
Divergent, with $\lim_{n\rightarrow \infty} a_n = \infty$ .

A sequence that converges to $1.7$ is, for example, $a_n=1.7-\frac{1}{n}$ $(n=1,2,3,...)$ .

As a series $(s_n)$ is also a sequence, it can also be convergent or divergent. The limit $s_{\infty}$ of a series, if it exists, has an interesting interpretation. It is an infinite sum, that is, the sum of infinitely many numbers:

Definition 3: Infinite sum a_1+a_2+...

Let $(s_n)$ be the series of a sequences $(a_n)=a_1, a_2, ...\,$ . We thus have

\begin{array}{lll} s_1 &=& a_1\\ s_2 &=& a_1+a_2\\ s_3&=& a_1+a_2+a_3\\ ...\\ \end{array}

We can think of the limit $\lim_{n\rightarrow \infty} s_n= s_\infty$ as

s_\infty = \underbrace{a_1+a_2+a_3+...}_{\text{infinitely many terms}}

Of course, as for any sequence, the limit or infinite sum can not exist, may be $\pm\infty$ or actually be a real number.

Extra: Convergent sequences - a deeper look for the brave

Show

What exactly do we mean by $a_n$ converges towards a real number $a_\infty$ ? Intuitively, it is reasonably clear: $a_\infty$ is that number towards which the terms of the sequence move if we move to the right in the sequence. However, note that none of the terms $a_n$ has to actually be the value $a_\infty$ (see figure below):

Let's describe this idea in mathematical terms:

Definition 4

The sequence $(a_n)$ converges to $a_\infty$ if for every small $\epsilon>0$ there exists a position $n_0$ in the sequence such that all subsequent terms $a_n$ are closer than $\epsilon$ to $a_\infty$ :

(a_\infty-\epsilon) \leq a_n \leq (a_\infty+\epsilon) \text{ for all } n\geq n_0

Side note: the above inequality is also often written like this:

\vert a_\infty-a_n\vert \leq \epsilon \text{ for all } n\geq n_0

Example 2

Intuitively, it is clear that the sequence

(b_n)=(1, 0.5, 0.25, 0.125, ...)

tends towards $0$ :

\lim_{n\rightarrow \infty} b_n = 0

Therefore, for any value $\epsilon>0$ , no matter how small, there must be a position $n_0$ in sequence so that for any position $n$ to the right of $n_0$ (that is, $n \geq n_0$ ) it is

0-\epsilon \leq b_n \leq 0+\epsilon

Determine such a $n_0$ for $\epsilon=0.00001$ .

Solution

It is $b_n=\left(\frac{1}{2}\right)^{n-1}$ (für $n=1,2,3,...$ ). As $b_n>0$ for all $n$ , find an $n_0$ with

b_n\leq 0+\epsilon \text{ for all } n \geq n_0

that is

=\left(\frac{1}{2}\right)^{n-1}\leq 0.00001

Let's find $n$ such that

\begin{array}{rll} \left(\frac{1}{2}\right)^{n-1} &=& 0.00001 \quad\vert \ln(..)\\ \ln\left(\left(\frac{1}{2}\right)^{n-1}\right) &=& \ln(0.00001)\quad \vert \ln(.)\\ (n-1)\ln(\frac{1}{2}) &=& \ln(0.00001) \quad \vert :\ln(\frac{1}{2}), +1\\ n&=&\frac{\ln(0.00001)}{\ln(\frac{1}{2})}+1 = 17.61\\ \end{array}

For all $n\geq 18$ ist $-0.00001\leq b_n\leq 0.00001$ (that is $n_0=\underline{18}$ ).

Extra: Divergent sequences - a deeper look for the brave

Show

Divergent sequences do not approach a single number. This can mean that the sequence jumps back and forth between different numbers forever and ever (see figure below):

But it can also mean that the sequence tends towards $\infty$ or towards $-\infty$ . Again, it is intuitively clear what we mean by this: A sequence $(a_n)$ tends towards $\infty$ if the terms become larger and larger without ever reaching a limit, that is, without an upper bound. And a sequence converges towards $-\infty$ if the sequence terms become more and more negative without ever reaching a limit or lower bound (see Figure below).

To be more exact:

Definition 5

A sequence $(a_n)$ converges to $\infty$ if for any arbitrarily large number $N$ there exists a position $n_0$ in the sequence such that all subsequent terms $a_n$ are greater than $N$ :

a_n \geq N \text{ for all } n\geq n_0

A sequence $(a_n)$ converges to $-\infty$ if for any arbitrarily large negative number $N$ there exists a position $n_0$ in the sequence such that all subsequent terms $a_n$ are smaller than $N$ :

a_n \leq N \text{ for all } n\geq n_0

Example 3

The sequence of square numbers

(a_n)=(1,4,9,16,...)

is divergent and converges to $\infty$ . Thus, for any arbitrarily large value $N$ , we can find a $n_0$ such that

a_n\geq N \text{ for all } n\geq n_0

Find a $n_0$ for $N=1\,000\,000$ .

Solution

It is $a_n=n^2$ . We have to find an $n_0$ such that

n^2 \geq 1\,000\,000 \text{ für alle } n\geq n_0

This is $n_0=\sqrt{1\,000\,000}=\underline{1\,000}$ .