Today we talked about infinite series which is an infinitely long sum of numbers. We followed the explanation in the book very closely, so I would recommend reading Section 10.1 - The Geometric Series.
Before doing any calculations, we started with three example infinite series:
Zeno’s series \(\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8} + \frac{1}{16} + \ldots = 1\).
Grandi’s series \(\displaystyle 1 + (-1) + 1 + (-1) + 1 + (-1) + \ldots = DNE\).
Harmonic series \(\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots = \infty\).
Then we talked about geometric series, which have the formula:
\[a + ax + ax^2 + ax^3 + ax^4 + \ldots = \frac{a}{1-x}\]
as long as the common ratio \(x\) satisfies \(|x| < 1\). Geometric series diverge if \(|x| \ge 1\), which means the infinite sum does not make sense.
We did the following examples in class:
What is the sum of \(\displaystyle 1+ \frac{2}{3}+\frac{4}{9}+\frac{8}{27} + \frac{16}{81} + \ldots\)?
If the sides of the outer square are both 2 meters long, find the total area of the darker squares:
Convert the repeating decimal \(0.36363636\ldots\) to a reduced fraction.
If \(f(x) = 1 + x + x^2 + x^3 + x^4 + \ldots\), what is \(f(-x^2)\)?
What is \((1+x+x^2+x^3+\ldots)(1+x+x^2+x^3+\ldots)\)?
What is the derivative of \(1+x+x^2+x^3+\ldots\) (both as a series and as a function)?
What is the integral of \(1-x^2+x^4-x^6+x^8 - \ldots\)?
What is the integral above when \(x=1\)?
Today we talked about what it means for an infinite series to converge. A series converges if the limit of its partial sums is a finite number. For a geometric series \(1+x+x^2 + x^3 + \ldots\), this happens if and only if \(|x| < 1\).
Then we talked about converting series that are written in term-by-term notation (like \(\tfrac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\)) into summation notation with a \(\Sigma\)-symbol. There are certain common patterns to watch out for:
Arithmetic patterns When you add the same amount each step. A formula for this pattern is \[a + bn\] where \(a\) is the first term and \(b\) is the step size.
Geometric patterns When you multiply by the same amount each step.
\[a r^n\] where \(a\) is the first term and \(r\) is the common ratio (the common amount you multiply by).
Alternating patterns If the terms alternate between positive and negative, you can write a formula for this by including a factor of \((-1)^n\).
We did the following examples (since these aren’t in most textbooks, I’ve included solutions too.).
Solution. This series has alternating terms and it has an arithmetic pattern in the denominator. To make the terms alternate, multiply by \((-1)^n\). The numerator is always 4, and the denominator starts at 1 and increases by 2 every step, so a formula for that arithmetic pattern would be \((1+2n)\). Putting it all together, we get:
\[\sum_{n=0}^\infty \frac{(-1)^n \cdot 4}{(1+2n)}.\]
Solution. This time, both the numerator and denominator follow arithmetic patterns, so the solution is:
\[\sum_{n=0}^\infty \frac{(5+2n)}{(3+n)}.\]
Solution. This time the denominator has a geometric pattern: each step multiplies by 2. So the solution is \[\sum_{n=0}^\infty \frac{1}{2(2^n)}\]
This solution could also be written as \(\sum_{n=0}^\infty \frac{1}{2^{n+1}}\) or as \(\sum_{n=1}^\infty \frac{1}{2^{n}}\). It is common to switch the starting point of a sum if it makes the formula simpler.
Solution. Here the pattern is alternating, and the numbers on the top are perfect squares. So the pattern on top looks like \((-1)^n n^2\), but that doesn’t come out quite right. If the sum starts at \(n=0\), you need \((-1)^n (n+1)^2\) to get the top right. The pattern on the bottom is geometric with common ratio 3, so the solution is: \[\sum_{n=0}^\infty \frac{(-1)^n (n+1)^2}{3^n}.\]
Another pattern that is common with series is the factorial. Some important things to know include the fact that \(0! = 1\), and that expressions with factorials in both the numerator and denominator can be simplified by cancelling the factors inside. For example, simplify the following expressions (without a calculator):
\(\displaystyle \frac{10!}{7!}\).
\(\displaystyle \frac{5! \, 4!}{(3!)^2}\).
\(\displaystyle \frac{(3n)!}{(3n+1)!}\).
Here is a simple idea that is pretty obvious.
Idea. An infinite sum cannot converge unless the terms in the sum are getting closer and closer to zero.
That’s one reason that Grandi’s series (\(1-1+1-1+1-1+1-\ldots\)) cannot converge. This simple idea is called the divergence test. Unfortunately, you can’t turn it around and say that a series with terms getting closer & closer to zero will converge. That isn’t true. An important set of examples are the p-series. A p-series is a series of the form:
\[\sum_{n = 1}^\infty \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \ldots.\]
We used integral calculus to show that these series converge if \(p > 1\) and they diverge otherwise. This is called the p-test for convergence. The special case when \(p=1\) is called the Harmonic Series and it is an important example to show that a series can diverge even if the terms approach zero. We finished with the following example:
Solution. First we converted the series to \(\Sigma\)-notation. We got \[\displaystyle \sum_{n=0}^\infty \frac{2+2n}{3^n}.\]
To find a larger series, we can either make the numerator larger or the denominator smaller. A good option here to to notice that \(2^n\) is bigger than \(2+2n\) for all \(n\) bigger than 1. So \[\sum_{n=0}^\infty \frac{2+2n}{3^n} \le \sum_{n=0}^\infty \frac{2^n}{3^n}.\] The series on the right-side is geometric with common ratio \(x = \frac{2}{3}\), so it converges. Therefore the smaller series on the left-side must also converge. This is an example of the comparison test which works the same for infinite series as it does for integrals.
Today we talked about the four types of special series: Geometric Series, p-Series, Alternating Series, and Power Series and how to tell if they converge or diverge.
A geometric series \(\displaystyle \sum_{n = 0}^\infty a r^n\) converges to \(\dfrac{a}{1-r}\) if \(|r| < 1\) and diverges otherwise.
A p-series \(\displaystyle \sum_{n = 1}^\infty \frac{1}{n^p}\) converges if \(p> 1\) and diverges otherwise. Unlike geometric series, there isn’t a nice formula for the sum of a p-series.
These are series with terms that alternate between positive and negative. They can be expressed as
\[\sum_{n=0}^\infty (-1)^n b_n\] where \(b_n\) is always positive.
Alternating series will converge if the following conditions are satisfied:
These conditions are called the alternating series test. I forgot to mention this in class (we’ll discuss it next week), but if an alternating series does converge, then you can estimate the error in a partial sum using the formula:
\[\text{Error} = |S-S_n| \le b_{n+1}\]
A good example of an alternative series is the alternating harmonic series: \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots\). Unlike the regular harmonic series which diverges, the alternative harmonic series converges by the alternating series test.
A power series is like an infinite degree polynomial. A power series has the form \[\sum_{n=0}^\infty a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\] All power series have a radius of convergence \(R\). When \(|x| < R\), the series converges, and when \(|x| > R\), the series diverges. At the endpoints, \(x = \pm R\), the series might converge or diverge. To find the radius of convergence, use the ratio test formula: \[R = \lim_{n \rightarrow \infty} \frac{|a_n|}{|a_{n+1}|}.\]
We did the following examples in class. For each example, you should be able to identify whether the series is alternating, geometric, a p-series, or a power series, and you should be able to determine whether it converges or diverges.
For each of the following, write out the first 4 terms of the alternating series, and determine whether it converges or diverges.
\(\sum_{n=0}^\infty \frac{ (-1)^n }{n!}\)
\(\sum_{n=0}^\infty (-1)^n \sqrt{n}\)
\(\sum_{n=0}^\infty (-1)^n \frac{4n}{3n+1}\)
Find the interval of convergence for the power series \(1 + x + x^2 + x^3 + \ldots\).
Find the interval of convergence for the power series \(\displaystyle \sum_{n = 0}^\infty \frac{(x-3)^n}{5^n}\).
Find the interval of convergence for the power series \(\displaystyle \sum_{n = 0}^\infty \frac{x^n}{n!}\).