Sum to Infinity

Calculate the sum to infinity of a geometric sequence.

When a geometric sequence has a ratio between -1 and 1, the terms diminish to zero. The sequence of partial sums converges to a limit, called the ‘sum to infinity’.

$S_{\infty}=\lim_{n\rightarrow\infty}\sum_{m=1}^{n}t_m=t_1+t_2+t_3+\dots+t_n$

When a geometric sequence has a common ratio $-1<r<1$ we can sum inifinitely many terms with this formula:

$S_{\infty}=\dfrac{a}{1-r}$

where $a$ is the first term and $r$ is the common ratio.

Example

A geometric sequence begins $40, 32, 25.6, 20.48 \dots$ . Calculate the sum to infinity of this sequence.

In this case, $a=40$ , $r=\dfrac{32}{40}=0.8$ . Therefore,

$S_{\infty}=\dfrac{a}{1-r}=\dfrac{40}{1-0.8}=\dfrac{40}{0.2}=200$

To calculate the sum to infinity of any sequence, the terms of the sequence must converge to zero. Not every sequence that converges to zero as a ‘sum to infinity’. This is called a necessary but not sufficient condition: not all sequences that converge to zero have a defined sum to infinityhttps://www.traditionrolex.com/33. See the harmonic series for an example of a sequence that converges to zero but the sequence of partial sums does not converge to zero. Nonetheless, the sum of any geometric sequences with $\abs{r}<1$ does have a limit.

Sum to Infinity of a Geometric Sequence

A geometric sequence has the form:

$a, ar, ar^2, ar^3, \dots, ar^n, \dots$

When $-1<r<1$ and $r\ne0$ , then the sequence converges to zero, regardless of the first term (Although $a=0$ doesn’t generate a very interesting sequence).

The sum formula

We already know how to add a defined number of terms of a geometric sequence:

$S_n&=\dfrac{a(r^n-1)}{r-1}$

where $a$ is the first term of the sequence, $r$ is the common ratio and $n$ is the number of terms to be added. This is a partial sum of the sequence.

The sum to infinity is defined when the sequence of partial sums converges.

The sum formula can be rearranged:

$\begin{align*}S_n&=\dfrac{a(r^n-1)}{r-1}\times\dfrac{-1}{-1}\\[10pt]&=\dfrac{-a(r^n-1)}{-(r-1)}\\[10pt]&=\dfrac{a(1-r^n)}{1-r}\end{align}$

This version of the summation formula is easier to use when $r<1$ .

Explain the sum to infinity formula

If $\left|r\right|<1$ then as $n\rightarrow \infty$ , $r^n \rightarrow 0$

Take any number that is less than one. For example, $0.7$ .

On the applet below, increase the value of $n$ to observe that as $n$ increases, the value of $r^n$ decreases towards zero.

Applet link

This applet offers 15 decimal places, however, if you consider

$\left(\dfrac{7}{10}\right)^{100}$

we know that this number is not zero. Very, very small, but not zero.

We say ‘tends to’ when a value that is changing approaches another value, which may be a constant or a function. We use the notation ‘ $\rightarrow$ ‘ for ‘tends to’.

Therefore if $\left|r\right|<1$ , then as $n\rightarrow \infty$ , $r^n \rightarrow 0$ .

The line above reads ‘if the absolute value of $r$ is less than 1, then as $n$ tends to infinity, $r^n$ tends to zero.’

Another way to write this is:

$\text{if }\left|r\right|<1, \quad \lim_{n\rightarrow\infty}r^n=0$

the line above reads: ‘If $\left|r\right|<1$ , the limit of $r^n$ as $n$ tends to infinity, is zero’.

Explained Example

When adding together an infinite number of terms of a geometric sequence that has $\left|r\right|<1$ , it would seem that there is a limit.

Consider the sequence $20, 10, 5, 2.5 \dots$

Here is the sum of the first several terms:

$\sum_{n=1}^{10}20\left(\dfrac{1}{2}\right)^{10}=\dfrac{20(1-0.5^{10})}{1-0.5}=39.96093\dots$

$\sum_{n=1}^{20}20\left(\dfrac{1}{2}\right)^{20}=\dfrac{20(1-0.5^{20})}{1-0.5}=39.9999618\dots$

$\sum_{n=1}^{30}20\left(\dfrac{1}{2}\right)^{30}=\dfrac{20(1-0.5^{30})}{1-0.5}=39.99999996\dots$

It would appear that as $n$ gets larger, $S_n$ gets closer to 40.

According to the summation formula for a geometric sequence, the sum of the first $n$ terms is:

$\dfrac{20(1-0.5^{n})}{1-0.5}$

However, as $n$ approaches infinity, $0.5^n$ approaches zero.

Therefore, the limit of this sum is

$\dfrac{20(1-0)}{1-0.5}=\dfrac{20}{0.5}=40$

We say

$\sum_{n=1}^{\infty}20\left(\dfrac{1}{2}\right)^{n-1}=40$

Conclusion

In general, because $r^n\rightarrow 0$ as $n \rightarrow \infty$ , we replace $r^n$ with zero in the formula:

$\text{if }\left|r\right|<1, \quad \sum_{n=1}^{\infty}a(r)^n=\dfrac{a(1-r^{n})}{1-r}=\dfrac{a(1-0)}{(1-r)}=\dfrac{a}{1-r}$

That is:

$\text{if }\left|r\right|<1, \quad \sum_{n=1}^{\infty}a(r)^{n-1}=\dfrac{a}{1-r}$

Practice

Try these problems on the CK12 website on summing infinitely many terms of a geometric sequence.

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