Wednesday 1 November class

Sorry this is delayed. I’m getting persistent “formula does not parse” messages and I cannot see why. So I’m posting as is. You can probably figure out what belongs in those places.

Topics:

• Introduction to sequences – was done last time with the sub. I added a couple of theorems and definitions which are listed below.

 

• Infinite series (introduction): partial sums, convergence, and a few examples.

Definitions I mentioned:

• A sequence is monotonic increasing if its terms never decrease: they always either increase or stay the same. A sequence is monotonic decreasing if its terms never increase: they always either decrease or stay the same.

If the terms always increase, we say the sequence is  strictly increasing.

If the terms always decrease, we say the sequence is  strictly decreasing.

• A sequence is bounded If there are two numbers M and m such that m < a_{n} < M for all n. Otherwise it is unbounded.

• A sequence is bounded from below If there is a number m such that $latex  a_{n}> m$ for all n.

• A sequence is bounded from above If there is a number M such that $latex  a_{n} < M$ for all n.

We won’t be using those last two right away: but keep them in the back of your mind. Obviously, a sequence is bounded if and only if it is bounded from below and from above.

 

The theorems about sequences that I mentioned:

• Absolute convergence: for a sequence \left\{a_{n}\right\}, if the sequence of absolute values \left\{\left|a_{n}\right|\right\} converges, then the original sequence \left\{a_{n}\right\} also converges.

This can be useful in determining whether or not an alternating sequence (one in which the signs of the terms alternate) converges.

• A convergent sequence is bounded: If the sequence \left\{a_{n}\right\} converges, then it is bounded.

(So if a sequence is unbounded, it cannot converge!)

• A bounded monotonic sequence converges: If the sequence \left\{a_{n}\right\} is bounded and monotonic (increasing or decreasing), then the sequence converges.

• Definition of a series (infinite series): given a sequence \left\{a_{n}\right\}, we form the infinite sum \displaystyle\sum_{i=1}^{\infty}a_{n}. This infinite sum is called a series.

Note: the index does not have to start at 1. At times it is more convenient to start at 0 or some other number.

• Partial sums: The n-th partial sum is S_{n} = \displaystyle\sum_{i=1}^{\infty}a_{n}

Again, the index may not start at 1. We start it wherever the sequence naturally starts.

• Convergence for a series: A series \displaystyle\sum_{i=1}^{\infty}a_{n} converges if the sequence of partial sums \left\{S_{n}\right\} converges: otherwise, it diverges.

Remember that when we say a limit is \infty or -\infty, it is just a way of saying that the limit does not exist in a particular way. In these cases we will say that the sequences diverges to infinity or diverges to negative infinity.

Examples I used in class:

We started out with the example on the top of p. 411. This is a very important example and will show you a lot about series, especially if you don’t already know (or have forgotten) that there is a formula for the sum. It is well worth examining in detail.

Examples:

• For the sequence a_{n} = n^{2}, n=1, 2 3, …, does the series converge or diverge?

We look at the first few terms of this sequence:

1, 4, 9, 16, 25, …

So the first few partial sums are

S_{1} = 1

 

S_{2} = 1+4 = 5

 

S_{3} = 1+4+9 = 14

 

 

And so on. It’s pretty obvious that this is an unbounded sequence, so it diverges. Therefore the series \displaystyle\sum_{i=1}^{\infty}n^{2} diverges.

Note: there is a formula you can write down for the partial sums: it is the formula for the sum of the first n squares. It is a useful thing to know (or at least, to know it exists!) Here is a nice discussion of several similar formulas, including adding the first 100 whole numbers, which supposedly Karl Gauss did as a youngster.

 

• For the sequence For the sequence a_{n} = (-1)^{n+1}, n=1, 2 3, …, does the series converge or diverge?

This is an alternating series: the first few terms are 1, -1, 1, -1, 1, -1, …

The first few partial sums are

S_{1} = 1

 

S_{2} = 1+(-1) = 0

 

S_{3} = 1+(-1)+1 = 1

 

 

And so forth. The partial sums are clearly not converging to anything. This series \displaystyle\sum_{i=1}^{\infty}(-1)^{n+1} diverges.

Note: in the previous example, the series diverges to infinity. Here, the series just plain diverges.

 

• Consider the sequence 9, 3, 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27},…

We can find a formula for the n-th term: observe that this sequence starts with 9 and then to get to the next term we multiply by \frac{1}{3} each time. (This is an example of a geometric sequence, as is the first example from the textbook: but please don’t go looking up formulas yet! Think it through!)

So the first few terms have these forms:

a_{1} = 9

 

a_{2} = 9\left(\frac{1}{3}\right)

 

a_{3} = 9\left(\frac{1}{3}\right)^{2}

 

a_{4} = 9\left(\frac{1}{3}\right)^{3}

 

a_{5} = 9\left(\frac{1}{3}\right)^{4}

 

a_{6} = 9\left(\frac{1}{3}\right)^{5}

 

And so on, so we see that

a_{n} = 9\left(\frac{1}{3}\right)^{n-1}

 

We computed the first few partial sums in class, enough to see that although they were growing, the growth was slowing down. Since the partial sums form a monotonic (increasing) sequence, if you can show they are bounded then you have proved that they converge. This is a little harder, but it’s possible to get a formula for the n-th partial sum in a similar way to the way we got it for the first example (the one from the textbook p. 411). I leave this to you as a challenge. The trick here is that when you form the partial sums, you should factor out the 9 and write the fraction factor as a mixed number as follows:

S_{1} = 9

 

S_{2} = 9 + 9\left(\frac{1}{3}\right) = 9\left(1+\frac{1}{3}\right) = 9\left(1\frac{1}{3}\right)

 

S_{3} =9\left(1\frac{1}{3}\right) + 9\left(\frac{1}{3}\right)^{2} = 9\left(1\frac{1}{3}+\frac{1}{9}\right) = 9\left(1\frac{4}{9}\right)

 

S_{4} =9\left(1\frac{4}{9}\right) + 9\left(\frac{1}{3}\right)^{3} = 9\left(1\frac{4}{9}+\frac{1}{27}\right) = 9\left(1\frac{13}{27}\right)

 

S_{5} =9\left(1\frac{13}{27}\right) + 9\left(\frac{1}{3}\right)^{4} = 9\left(1\frac{13}{27}+\frac{1}{81}\right) = 9\left(1\frac{40}{81}\right)

 

If you look closely at how we get each new numerator, the numerators are the sums of powers of 3:

3^{0} = 1 and the first numerator is 1

3^{1} = 3 and the second numerator is 1+3 =4

3^{2} = 9 and the third numerator is 1+3+9 =13

3^{3} = 27 and the fourth numerator is 1+3+9+27 = 40

and so forth.

Thus the fraction in S_{n} always has denominator 3^{n-1} and its numerator is the sum of the first n-1 squares of 3. There is a formula for that! You can get it by a careful application of the formula for the first n perfect squares. (If you look at the blog post I linked above, they work out something similar for the sum of the first n powers of 2). That’s the challenge! and then you can find the limit of the S_{n}, which does exist: this series converges.

To be continued!

Homework:

• Review the examples, definitions and theorems discussed in class. (Not all of the examples are written up here, alas.)

• Do the WeBWorK on sequences: due by Sunday evening:do not wait to the last minute!

• I am also adding a short WeBWorK on series, due Sunday.

• Start working on the Test 2 redo problems if you choose to do them. Please pay careful attention to the instructions. In particular, the due date/time is firm.

 

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!