Monday 20 November class

Topics:

• Discussion of some homework problems, including estimating the sum of a convergent alternating series

• The Root test for absolute convergence

• Power series: definitions and vocabulary

A power series centered at a is a series which has the form \displaystyle\sum_{n=0}^{\infty}c_{n}(x-a)^{n}, where the c_{n} are real numbers called the coefficients of the series. Note that (as usual) the sum does not have to start at n=0. It starts wherever it starts. (Note also the similarity to Taylor polynomials!)

The series may or may not converge for various values of the variable x. If the series converges at all, there is a non-negative number R (possibly infinity) such that:

If |x-a| < R the series converges absolutely,

If |x-a|>R the series diverges.

R is called the radius of convergence of the power series.

Note that if |x-a| = R, the series may either converge or diverge. We have to look more closely to see what happens then.

This all should remind you of what happens when we use the Ratio (or Root) tests for absolute convergence. The Ratio Test will be our main tool for analyzing power series and their convergence. (We will have to use all of our other tests to determine what happens at the endpoints where the limit is 1 from that test, though.)

 

Example: \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{4^{n}}(x+3)^{n}

This is a power series centered at -3. Let’s find the radius of convergence using the Ratio Test!

We can forget about the powers of -1, since we are taking absolute values

\left|\frac{a_{n+1}}{a_{n}}\right| = \frac{\frac{n+1}{4^{n+1}}|(x+3)^{n+1}|}{\frac{n}{4^{n}}|(x+3)^{n}|} = \frac{n+1}{4n}|x+3|

 

\displaystyle\lim_{n\rightarrow\infty}\frac{n+1}{4n}|x+3| = \frac{1}{4}|x+3|

So the power series will converge (absolutely) if $latex \frac{1}{4}|x+3| <1$, in other words it will converge absolutely if |x+3| <4, and it will diverge if |x+3| >4: so the radius of converge is 4, according to the definition given above.

We would have to to look more closely using other tools to find out whether or not the series converges when |x+3| =4. We will do that next time.

 

 

 

 

 

Some notes from class:

Here is a version of problem 11 from the WeBWorK Ratio and Root Tests, which was partially worked out in class.

The problem was to use the root test on the series \displaystyle\sum_{n=1}^{\infty}\left(1+\frac{1}{5n}\right)^{-n^{2}} , which resulted in our having to find the limit

\displaystyle \lim_{n\rightarrow\infty}\left(\left(1 + \frac{1}{5n}\right)^{-n^{2}}\right)^{\frac{1}{n}} = \lim_{n\rightarrow\infty}\left(1 + \frac{1}{5n}\right)^{-n}

I rewrote that to simplify it somewhat:

\displaystyle \lim_{n\rightarrow\infty}\left(1 + \frac{1}{5n}\right)^{-n} =\lim_{n\rightarrow\infty}\left(\frac{5n+1}{5n}\right)^{-n} = \lim_{n\rightarrow\infty}\left(\frac{5n}{5n+1}\right)^{n}

You will need to use L’Hopital’s Rule to find that limit, since it has the indefinite form 1^{\infty}

Here is the basic procedure:

Let y = \displaystyle\lim_{n\rightarrow\infty}\left(\frac{5n}{5n+1}\right)^{n}.

To change this problem into one that will have a simpler indefinite type, take the logarithm:

\ln(y) = \displaystyle\ln\left(\lim_{n\rightarrow\infty}\left(\frac{5n}{5n+1}\right)^{n}\right) =\lim_{n\rightarrow\infty}\ln\left(\frac{5n}{5n+1}\right)^{n} since the logarithm is a continuous function.

You would then use the “log of a power” rule to turn this into a limit which has the form \infty\cdot 0, and then do a rewrite to get the form \frac{0}{0}. I have given up on trying to type in the rest of the procedure here, as WordPress will not display it correctly. Please see the note posted on Piazza for the rest of this.

If you want to see an example worked out for the indefinite form $latex 1^{\infty}$, see Example 192(1) on p. 329 of the textbook. (This is the limit which is sometimes used to define e, and you may be able to find it online as well, but why bother since it is in your text?)

One of the problems in that WeBWorK involved factorials. For simplifying quotients of factorials, you just have to remember how they are defined. Here is the example I discussed:

\frac{(2n+1)!}{(2n+3)!} = \frac{(2n+1)!}{(2n+3)(2n+2)(2n+1)\cdots(1)} = \frac{(2n+1)!}{(2n+3)(2n+2)(2n+1)!} =\frac{1}{(2n+3)(2n+2)}

When you have to simplify a quotient of factorials, always expand them out (at least in your mind) to see how they can be reduced.

 

Homework:

• Review the examples and definitions discussed in class. Make sure that you understand what a power series and the radius of convergence for a power series are.

• Finish the WeBWorK “Ratio and Root Tests”. You do not need to work on the Power Series WeBWorK yet: you can look at it if you like.

• Here is a video from PatrickJMT showing how to find the interval of convergence for a power series in the simplest case, where the center is at 0. The new thing is how to determine whether or not the endpoints of the interval (where the limit of the ratio test is 1) are to be included in the interval of convergence or not. We have to use one of the earlier tests about convergence of series to figure this out separately for each endpoint. We will go into this in more depth next time. Also, on to Taylor Series.

• There will be a quiz next time. The topics will be the alternating series test, and the ratio and root tests.

• Don’t forget to fill out the Post-Test Strategic thinking survey: research has shown that using this type of survey and thinking about your use of resources significantly improves test scores!

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!