Limit of arithmetic and geometric sequences and series

Since the arithmetic sequence $(a_n)$ always grows or falls by the same amount $d$ , it is clear that such a sequence tends towards plus infinity or minus infinity (i.e. diverges). For example, for $d=3$ and $d=-3$ and $a_1=1$ we have the sequences

(a_n)=(1, 4, 7, 10, ...)

(a_n)=(1,-2, -5, -8, ...)

The exception is the trivial case $d=0$ , which is rather uninteresting:

(a_n)=(1,1,1,1,...)

Indeed, this sequence converges to $1$ . Let's summarize:

Theorem 1

For an arithmetic sequence $(a_n)$ with common difference $d$ the following holds true

\lim_{n\rightarrow \infty} a_n = \begin{cases} \infty \text{ divergent}, & d>0\\-\infty \text{ divergent}, & d<0\\a_1, & d=0 \end{cases}

It is also clear that the terms $s_n$ tend towards $\infty$ or $-\infty$ . Or formulated differently, for the infinite sum of an arithmetic sequence it is

s_\infty=a_1+a_2+... = \pm \infty

So the convergence behaviour of arithmetic sequences and their sums is not very interesting. What about the geometric sequence? This depends on the common quotient $q$ . Let's look at the following examples, where we always set $a_1=1$ , so we have geometric sequences of the form

(a_n)=(1, q, q^2, q^3, ...)

$q$ is between $-1$ and $1$ ( $-1<q<1$ ): e.g. for $q=\frac{1}{2}$ we have $(a_n)=(1, 0.5, 0.25, 0.125, ...)\xrightarrow[]{n\rightarrow \infty} 0$ and we see that the sequence converges to $0$ . The same is true for $q=-\frac{1}{2}$ : $(a_n)=(1, -0.5, 0.25, -0.125, ...)\xrightarrow[]{n\rightarrow \infty} 0$ And the same is true for any other $q$ with $-1<q<1$ . This is true even for $q=0.99999999$ , but in this case the sequence tends very slowly towards $0$ . This can easily be checked with the calculator.
$q$ is greater than $1$ or less than $-1$ ( $q>1$ or $q<-1$ ): e.g. for $q=2$ we have $(a_n)=(1,2,4,8,...)$ and tends to infinity (diverges). This is also the case for smaller $q$ , such as $q=1.0000000000001$ . The sequence then grows much more slowly, but still tends towards infinity, as can easily be checked with the calculator. If we start with a negative $a_1$ , it is easy to see that the sequence tends to $-\infty$ . For instance, for $a_1=-1$ we get $(a_n)=(-1,-2,-4,-8,...)$ (diverges). For $q=-2$ we get the sequence $(a_n)=(1,-2,4,-8,...)$ and we see that the sequence diverges.
for $q=1$ and $q=-1$ we get the uninteresting sequences $(a_n)=(1,1,1,1,...)$ which converges to $1$ and $(a_n)=(1,-1,1,-1,...)$ which is divergent.

Let us summarise:

Theorem 2

For a geometric sequence $(a_n)$ with common quotient $q$ holds

\lim_{n\rightarrow \infty} a_n = a_1 q^{n-1}= \begin{cases} \pm \infty \text{ divergent} & (q>1)\\ a_1 & (q=1)\\ 0 & (-1 < q < 1)\\ \text{divergent} & (q\leq -1) \end{cases}

For $-1<q<1$ , the geometric series $(s_n)$ converges as well. Indeed, because $q^n$ converges to $0$ in this case, we see that the sum formula

s_n = a_1+a_2+...+a_n = a_1\frac{1-q^n}{1-q}

converges to

s_\infty = a_1+a_2+... = a_1\frac{1}{1-q}

Example 1

Find the infinite sum of the sequence $(a_n)=-\frac{1}{4},\frac{1}{6},-\frac{1}{9}, \frac{2}{27},...$

Solution

This is a geometric sequence with $a_1=-1/4$ and $q=-2/3$ . Because $-1<-2/3<1$ , it follows

\begin{array}{lll} s_\infty&=&-\frac{1}{4}+\frac{1}{6}-\frac{1}{9}+\frac{2}{27}- ...\\ &=&-\frac{1}{4}\cdot \frac{1}{1-\left( -\frac{2}{3} \right)}\\ &=&\underline{-\frac{3}{20}}\end{array}

Example 2

Determine the infinite sum $1+0.5+0.25+0.125+...$

Solution

The sequence $(a_n)=(1,0.5, 0.25, 0.125,...)$ forms a geometric sequence with $a_1=1$ and $q=0.5$ . As $-1<0.5<1$ it is

\begin{array}{lll} s_\infty &=&1+0.5+0.25+0.125+...\\ &=&1\cdot \frac{1}{1-0.5}\\ &=&\underline{2}\end{array}

Warning

The summation formula $s_\infty = a_1\frac{1}{1-q}$ applies only if $-1<q<1$ . If $q$ is not between $-1$ and $1$ we could in principle still use this formula to calculate a number, but this number is then not the infinite sum (which actually is $\pm \infty$ or does not exist). For example, the geometric sequence with $a_1=1$ and $q=2$ is

(a_n)=(1,2,4,8,...)

and it is quickly apparent that

s_\infty=1+2+4+8+...=\infty

But if we falsely use the sum formula ( $2$ is not between $-1$ and $1$ ), we get

s_\infty = 1\cdot \frac{1}{1-2}=-1 \quad \text{FALSE!, not the infinite sum}