Sum to Infinity

The sum to infinity of a sequence is the sum of an infinite number of terms in the sequence.

    \[S_{\infty}=\sum_{n=1}^{\infty}t_n=t_1+t_2+t_3+\dots\]

It is only possible to compute this sum if the terms of a sequence converge to zero. Even then, it is not always possible. This is called a necessary but not sufficient condition – for a sum to infinity to be defined, the sequence must converge to zero – but not all sequences that converge to zero have a defined sum to infinity. We will see on this page that it is possible to compute this sum for all geometric sequences that have a common ratio -1<r<1.

For any sequence that diverges, the sum of the sequence also diverges.

The arithmetic sequence 4, 7, 10, 13, 16, … diverges

The series 4 + 7 + 10 + 13 + 16 also diverges.

Sum to Infinity of a Geometric Sequence

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). Let’s examine the sum to infinity of a couple of examples, then generalise. First, we need to be OK with the following two ideas:

Concept 1: Multiply a fraction by \frac{-1}{-1}

Notice that

    \[\dfrac{7-2}{7-3}=\dfrac{2-7}{3-7}\]

Both sides in the line above are equal to positive \frac{5}{4}.

When we multiply any value by 1 (the multiplicative identity) we do not change its value. The fraction \frac{-1}{-1}=1. Let’s begin with the left hand side and multiply by \frac{-1}{-1}.

    \begin{align*}&\dfrac{7-2}{7-3} \times \dfrac{-1}{-1}\\[10pt]=&\dfrac{-(7-2)}{-(7-3)}\\[10pt]=&\dfrac{-7+2}{-7+3}\\[10pt]=&\dfrac{2-7}{3-7}\end{align}

We do this in order to rearrange our summation formula for a geometric series as follows:

    \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.

Concept 2: 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’.

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:

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

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,

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

Practice

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

Next: Unit Assignment and Quiz