Convergence of Series

Our goal in this part of the course is to approximate complicated functions with much simpler ones, namely polynomials. We have seen one example where the function $ \frac{1}{1-x} $ could be expressed in terms of its Taylor series

\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \]

for values $ |x| < 1 $ . To interpret this statement, however, we need to understand when the sum of an infinite number of numbers is finite (in this case, we say the series converges). That is the goal of this section. Basically, we will aim to present some simple techniques for determining when a series will converge.


Geometric Series

We have already seen that a geometric series will converge precisely when the ratio of successive terms is smaller than one in absolute value:

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

converges when $ |r| < 1 $ . For instance, the series $ \sum_{n=0}^\infty (\frac 23)^n $ converges.


The harmonic series

The harmonic series is obtained by summing the reciprocals of the positive integers:

\[ \sum_{n=1}^\infty \frac 1n = 1 + \frac 12 + \frac 13 + \frac 14 + \ldots \]

You might think that this series converges because the addends are approaching zero. However, the issue of convergence is a little more subtle than that. We can study this series by comparing it to an integral.

In this diagram is shown the graph $ y = \frac 1x $ ; the area under this graph is shaded in orange. Also shown are rectangles built from the function. From the picture, it is clear that the area of the rectangles is greater than the area under the curve.

How much area is represented by the rectangles? The leftmost rectangle has height 1 and base 1 so its area is 1. The next rectangle has height $ \frac 12 $ and base 1 so its area is $ \frac12 $ . You can imagine that the n th rectangle will have height $ \frac 1n $ and base 1 giving an area of $ \frac 1n $ . Putting this all together, we see that the area of the rectangles is $ 1 + \frac 12 + \frac 13 + \ldots $ ; in other words, the area of the rectangles represents the harmonic series. Since this area is larger than the area under the graph, we have:

\[ 1 + \frac 12 + \frac 13 + \ldots > \int_1^\infty \frac 1x~dx = \ln x |_1^\infty = \infty \]

In other words, the area of the rectangles is infinite and so the harmonic series diverges.

More generally, we can compare series of the form $ \sum_{n=1}^\infty \frac{1}{n^p} $ to the integral $ \int_1^\infty \frac{1}{x^p}~dx $ in precisely the same way. This leads to the conclusion that

\[ \sum_{n=1}^\infty \frac{1}{n^p} ~~~~~~\big\{ \begin{array}{ll} \mbox{ converges if } & p > 1 \\ \mbox{ diverges if } & p \leq 1 \end{array} \]

For example, the series

\[ \sum_{n=1}^\infty \frac{1}{n^2} = 1 + \frac 14 + \frac 19 + \frac 1{16} + \ldots \]

converges. Notice, however, that comparison does not give us a value to which the sum converges.


Alternating series

We can modify the harmonic series to produce the alternative harmonic series :

\[ \sum_{n=1}^\infty (-1)^{n+1} \frac 1n = 1 - \frac 12 + \frac 13 - \frac 14 + \ldots \]

Notice that this looks a lot like the harmonic series only the sign alternates from term to term. In the demonstration below, we will study this example:

Whereas the harmonic series does not converge, the alternating harmonic series does converge due to the alternation of signs. In some sense, two consecutive terms nearly cancel one another out and this results in the series converging.

More generally, suppose that we have a series $ \sum_{n=0}^\infty a_n $ such that

  1. $ a_n $ and $ a_{n+1} $ have opposite signs; that is, the signs of the individual terms alternate.
  2. $ |a_{n+1}| < |a_n| $ ; that is, the terms are decreasing in absolute value.
  3. $ a_n \to 0 $ ; that is, the terms are approaching zero.

Then, in the same way as the alternating harmonic series, we can be sure that the series $ \sum_{n=0}^\infty a_n $ converges.


Factorials

Given an integer n, we define n factorial to be the product of the first n integers and we write $ n! = 1\cdot 2 \cdot 3\cdot \ldots n $ . By convention, we will take $ 0! = 1 $ . Notice that

\[ \begin{array}{c} 0! = 1 \\ 1! = 1 \\ 2! = 1\cdot 2 =2\\ 3! = 1\cdot 2\cdot 3 = 6 \\ 4! = 1\cdot 2 \cdot 3 \cdot 4 = 24 \\ 5! = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 \\ 6! = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 720 \end{array} \]

As you can see, factorials grow remarkably fast. In this example, we will consider the series

\[ \sum_{n=0}^\infty \frac{1}{n!} = 1 + 1 + \frac 12 + \frac 16 + \frac{1}{24} + \ldots \]

We can see that this series converges by noticing:

\[ \frac{1}{n!} = \frac 11 ~~\frac 12 ~~\frac 13 ~~\ldots ~~\frac 1n \leq 1 \frac 12~~\frac 12~~\ldots ~~\frac 12 = \frac{1}{2^{n-1}} \]

if $ n \geq 2 $ . In other words, the series we are interested in eventually becomes smaller than a geometric series which converges.

\begin{eqnarray*} \sum_{n=0}^\infty \frac{1}{n!} & = & 1 + 1 + \frac 12 + \frac 16 + \frac 1{24} + \ldots \\ & = & 1 + 1 + \frac 12 + \frac 14 + \frac 18 + \ldots \\ & = & 1 + (1 + \frac 12 + (\frac 12)^2 + (\frac 12)^3 + \ldots \\) & = & 1 + 2 = 3 \end{eqnarray*}

This is interesting because, in addition to showing that the series converges, it shows that it converges to a number smaller than 3. The following demonstration studies will show you the sum of the first few terms in this series:

Even more remarkably, this demonstration suggests that the series converges to the number $ e $ . We will soon see why this is. In the meantime, you may want to practice this idea by showing that the series

\[ \sum_{n=0}^\infty \frac{a^n}{n!} \]

converges no matter what value of $ a $ is used.