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. We did the following examples in class:
What is the sum of \(\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8} + \frac{1}{16} + \ldots\)?
What is the sum of \(\displaystyle 1+ \frac{2}{3}+\frac{4}{9}+\frac{8}{27} + \frac{16}{81} + \ldots\)?
What is \(a(1+x+x^2+x^3+x^4 + \ldots)\), both as an infinite series and as a function?
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 \(1+x+x^2+x^3+\ldots\)?
What happens if you substitute \(-x^2\) in place of \(x\) in the series \(1+x+x^2 + x^3 + \ldots\)?
What is the integral of the last series?
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 1, 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:
\[\sum_{n=0}^\infty \frac{(-1)^n}{(1+2n)}.\]
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.
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):
\(\displaystyle \frac{10!}{7!}\).
\(\displaystyle \frac{5! \, 4!}{(3!)^2}\).
\(\displaystyle \frac{(3n)!}{(3n+1)!}\).
We finished today by introducing p-Series which are 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 the integral test to show that these series converge if \(p > 1\) and they diverge otherwise. 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.
Today we talked about Alternating Series and Power 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:
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 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 indentify 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.
Alternating harmonic series \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots\)
\(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \ldots\)
Expressing the previous sum in summation notation, we get \(\sum_{n = 0}^\infty (-1)^n \frac{1}{2n+1}\). Estimate the error if you estimated this sum by using the partial sum up to \(n=1000\).
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 radius of convergence of the power series \(\sum_{n = 1}^\infty \frac{2^n x^n}{n^2}\).