Geometric Series

Check out my new book on sequences and series!

You can read it on your Kindle, iPad, or Android device.


Click to see Calculus Sequences and Series: Problems and Solutions

The first type of series we usually learn about is called a geometric series. They’re so named because the ratio or successive terms is constant, but who cares about that. What do they look like? Here are two examples:

    \[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \]

and

    \[ 3 + \frac{3}{5} + \frac{3}{25} + \frac{3}{125} + \dots\]

are both geometric series. In sum notation, we can write these series as

    \[ \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n \]

and

    \[ \sum_{n=0}^{\infty}3\left(\frac{1}{5}\right)^n \]

respectively.

Another example of a geometric series is

    \[ \sum_{i=0}^{\infty} 7\cdot 2^i = 3 + 7\cdot 2 + 7\cdot 2^2 + 7\cdot 2^3 + \dots \]

However, we know by the divergence test that this series doesn’t converge, so it’s not as interesting. (Does the divergence test really work on this series? Check it!)

In general, a geometric series is one whose terms are a constant times a constant raised to the power of the variable you’re summing over. They look like this:

    \[ \sum_{n=0}^{\infty} a r^n \]

where a and r are constants (numbers). Note that we could also write this series as

    \[ \sum_{n=1}^{\infty} ar^{n-1} \]

Geometric series are special. We can determine whether or not they converge, but we can also say exactly what their sum is. Usually this is very, very difficult. But not so with geometric series. We can use this observation to find the partial sums: since a_n = ar^{n-1} for a geometric series, s_n = a + ar + ar^2 + \dots + ar^{n-1}. Multiplying by r and subtracting, we get

    \begin{align*} s_n - rs_n & = a + ar + ar^2 + \dots + ar^{n-1} \\ &\quad - (ar + ar^2 + \dots + ar^{n-1} + ar^n)\\ & = a - ar^n. \end{align*}

Solving for s_n, we obtain

    \[ s_n = \frac{a-ar^n}{1-r}. \]

What happens to this sequence as n\rightarrow\infty? If |r|<1, r^n just gets smaller and smaller, approaching zero. So, when |r|<1, s_n\rightarrow \frac{a}{1-r}. On the other hand, when |r|\geq 1, the test for divergence shows that the series diverges. In summary, we have

    \[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \]

if |r|<1, and the series diverges otherwise.

Sometimes geometric series are hiding and we have to do some algebra to get them to show themselves. Consider this example:

    \[ \sum_{n=0}^{\infty}\frac{3^{n+1}}{2^{2n}}. \]

Doesn’t really look like a geometric series, quite. But we can rearrange the terms like so:

    \begin{align*} \frac{3^{n+1}}{2^{2n}} & = \frac{3\cdot 3^n}{(2^2)^n}\\ & = 3\frac{3^n}{4^n}\\ & = 3\left(\frac{3}{4}\right)^n. \end{align*}

So

    \[ \sum_{n=0}^{\infty}\frac{3^{n+1}}{2^{2n}} = \sum_{n=0}^{\infty} 3\left(\frac{3}{4}\right)^n, \]

and now we can tell that this is a geometric series. Furthermore, since 0<\frac{3}{4}<1, it converges to the value

    \[ \frac{3}{1-\frac{3}{4}} = 12. \]

Here are some more problems involving geometric series:

  1. The series

        \[ \sum_{n=1}^{\infty}\frac{1}{n^2} \]

    is not geometric, because n (the variable being summed over) doesn’t appear in the exponent of a constant. Similarly

        \[ \sum_{n=1}^{\infty}\frac{3}{n2^n} \]

    isn’t geometric because there’s an extra n (the one that isn’t in an exponent).

  2. The series

        \[ \sum_{i=1}^{\infty} \frac{1}{2^{i-1}} \]

    is geometric with a=1 and r=1/2. It converges because |r|<1, and its value is 2.

  3. The series

        \[ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \]

    is geometric. Here a=1/3, r=1/3, the series converges, and its value is 1/2.

  4. The series

        \[ \sum_{k=0}^{\infty} \frac{5^k}{4^{k+2}} \]

    is geometric. What are a and r? Does the series converge? If so, to what value?