From Euclid to Spigotry #13
Infinite Series
Last modified: September 15, 2004
Lecture topic 1. The "alternating series test" -- an informal approach
Imagine that you are doing research in Kerala on some newly-discovered Sanskrit manuscripts from the 16th century.  You translate one of them and find the following:
"The Maharajah's treasure is buried on the road that runs past this monastery.  Here is how to
find it.
1. Walk 4 miles east.
2. Then walk 4/3 miles west.
3. Then walk 4/5 miles east.
4. Then walk 4/7 miles west.
5. Then walk 4/9 miles east.
6. Then walk 4/11 miles west.
7. Then walk 4/13 miles east.
8. Then walk 4/15 miles west.
9. Continue this process until you are fairly sure of the location, then dig."

a. Draw a diagram showing that if you dig a trench between the locations that you reach on any two consecutive steps, you are sure to find the treasure.
b. Show that you can also find the treasure by stretching a rope around the local racecourse, which happens to be a circle whose diameter is 1 mile, then straightening out the rope along the road east of the monastery.
Lecture topic 2. The "alternating series test" -- a formal approach
Let An = a1 - b1 +a2 - b2 +...+an.
Let Bn = a1 - b1 +a2 - b2 +...+an - bn.
where the positive quantities ai and bi satisfy
a1 > b1 > a2 > b2 > a3 > b3 >  ... >an > bn
and you can find a value of bn that is as small as you like simply by choosing a large enough
value of n.
a. Prove that for all n > 1, An+1 < An and Bn+1 > Bn.  Then prove by induction that for any m,
An+m < An and Bn+m > Bn. This means that once you have calculated Bn and An for some n, all subsequent values will lie in between them.
b. Suppose that someone suspects that there are two different numbers x and y, with x < y, such that
Bn < x < An and Bn < y < An for every value of n.
Prove that this cannot happen.  Hint: you can make An - Bn less than y - x.

Lecture topic 3: The geometric series test - an informal approach
The gold rush is on in Kerala.  Hundreds of Harvard undergraduates have abandoned biology, economics and computer science to concentrate in Sanskrit, and it is not long before one of them
finds a manuscript that reads
"The Sultan's treasure is buried along the road that runs past the palace. Here is how to find it:
1. Walk 1 mile east.
2. Walk east 2/3 as far as in the previous step
3. Walk east 2/3 as far as in the previous step.
4. Keep doing this until you are fairly sure of the location, then dig."

a. Show that the location of the treasure is given by the sum of the infinite series
1 + x + x2 + x3 + ...., where x = 2/3 and that this sum equals 3
Hint: 1 + x + x2 + x3 + ...+ xn = 
b. Alas, the treasure is not there.  But then another manuscript comes to light.  It reveals that the
Sultan's metal measuring rods were forged during the hot season, but that the treasure was buried during the cool season, when the rods had shrunk somewhat.  As a result, the correct distances are all less than the first manuscript suggested, by an unknown factor that may change from one step to the next.  Argue that digging up 3 miles of road east of the palace is still certain to reveal the treasure.
Lecture topic 4: The geometric series test - a formal approach
Suppose that an infinite series of terms is of the form
A0 + A1 x + A2 x2 + ....
where 0< x < 1 and the positive quantities A0, A1, A2,...  all satisfy P < Ai < Q
a. Prove that the sum of the series lies between  and 
b. A famous series discovered by Euler is
+...
Show that this series has a sum that lies between 3/2 and 2.
c. By writing the series in the form
,
show that the sum lies between 14/9 and 5/3. (In fact, the sum is p/2)

Lecture topic 5: Conditionally convergent series:
By starting with the infinite series for , show that
ln(2) = 
Then show how to rearrange this conditionally convergent series so that its sum is .
This example is credited with waking up the world of mathematics to the need to define convergence carefully. 
Start with the geometric series
= 
Integrate both sides:
ln(1+x) = 
Set x = 1:
ln(2) = 
(This step requires justification in the form of Abel's limit theorem -- Knopp  p. 177)
This conditionally convergent series is just as famous as Gregory's series.
Now multiply both sides by 1/2:
=   
Add term-by-term, combining fractions with the same denominator.
= 
Think about it: all the terms with odd denominators show up from the original series.  All the terms with even denominators eventually show up as the sum of two negative terms whose denominators are divisible by 4.  So this series has the same sum as the original series, but in a different order.  Yet it has a different sum!
This sort of thing cannot happen with an absolutely convergent series: one that still converges even if you take the absolute value of each term.  So if you start with, for example,
,
which was summed by Euler (see Beckmann, page 153), there is no way that you can change the sum by reordering the terms.
For a generalization of this result see the "accountant's dream" program below.

Lecture topic 6 -- the "accountant's dream" program

Every accountant would love to be able to take a sequence of numbers and obtain any desired value as the sum.  For a finite sequence this cannot be done, but for conditionally convergent series it is possible to achieve any sum by rearranging the series.  This remarkable result is known as the Riemann Series Theorem.

Examples: #1  1 - 1/2 + 1/3 -1/4 + ...  (topic 5)

#2 1 - 1/3 + 1/5 - 1/7 + 1/9  (Gregory's series)

If the terms are summed in the specified order, #1 sums to ln(2) while #2 sums to p/4

To achieve a different sum s, proceed as follows:

Add positive terms, in order, until the sum exceeds s.
Add negative terms, in order, until the sum is less than s.
Repeat these two steps until the magnitude of the largest term that you used is smaller than
some predetermined small quantity e.  Your sum is now within e of s and will remain so.

You will compute a nest of intervals, all of which include s, whose length goes to 0 (i.e. becomes and remains less than e).   Thus the series converges to s.  Of course, you may be using the 10000th positive term before you use the 10th negative term!

A fairly small modification of any program that sums Gregory's series is all that is needed.
Once people see why the program works, they will understand the proof of the theorem.
Lecture topic 7: Taylor series
The Taylor series for a function f(x) about x = 0 is the power series
P(x) = f(0) + f'(0) x + 
Show that the difference between f(x) and its Taylor series P(x),
g(x) = f(x) - P(x) 
has the property that g(x) and all its derivatives are 0 at x=0.  (This is not quite the same as saying that g(x) is the zero function, which is why many calculus texts gloss over the proof of Taylor's theorem.)
As an example, choose f(x) = cos x, expand about x = 0, and show that
cos x = 1 - 
Next choose f(x) = sin x, expand about x = 0, and show that
sin x = x - 
For a final example, choose f(1+x) = (1+x)a
and show that f(1+x) = 1 + ax +  + ...
This is another way to prove the binomial theorem.

Lecture topic 8: Trig functions defined by power series:

Define 

The task is to show that these function, defined without reference to geometry, have all the key properties of the sine and cosine functions. But don't assume that!
a. Express S'(x) in terms of C(x) and C'(x) in terms of S(x).  Thereby obtain expressions for the first four derivatives of each function (and ultimately, as the answer comes out, expressions for all derivatives.)
b. Prove that  (Hint: first differentiate the formula)
c. Using Taylor expansions of S(a+x) and C(a+x) about x=0, show that
 S(a+b) = S(a)C(b) + C(a)S(b) and that C(a+b) = C(a)C(b) - S(a)S(b)
Lecture topic 9: Pi without geometry
Using the functions C(x) and S(x) that were introduced in the preceding topic,
define p by the property that 1 < p/2 < 2 and that C(p/2) = 0. To show that this specifies p/2 uniquely, you have to make the case that C(1)>0, C(2) < 0, and C'(x) is negative for 1 < x < 2.
Then the graph of C(x) crosses the x axis precisely once between 1 and 2.
Using the formulas proved in topic 8c, calculate S(p/2), C(p), S(p), C(2p), and S(2p).
Prove that C(x + 2p) = C(x) and that S(x + 2p) = S(x).