1002.html

Module 10 : Serious About Series

1002 : Convergence Tests

O B J E C T I V E

In this project we will examine and use some of the various tests that can be used to determine if a series converges or diverges.

S E T U P

In this project we will use the following command packages. Type and execute this line before begining the project below. If you re-enter the worksheet for this project, be sure to re-execute this statement before jumping to any point in the worksheet.

> restart; with(plots):

Warning, the name changecoords has been redefined

___________________________________________________________________________________

A. The Integral Test & P- Series

___________________________________________________________________________________

There are many occasions where a series is closely related to an integral. In these cases,the integral will converge or diverge if and only if the corresponding sum converges or diverges. This result is calle the Integral Test. Remember a sequence is function defined on the domain of natural numbers. If it makes sense to define the same function for all positive real numbers, and that functioin is integrable ( able to be intergrated) , then the integral test will apply. Lets look at an example.

Here is a function, the sequence it defines, and a plot of the sequence and function.

> f := x -> 1/x^(3/2);

> seq( evalf( f(k)), k = 1..12);

> plot( [f(x), f( floor(x)), f(floor(x+1))], x = 1..12, thickness = [4,2,2], color = [blue, coral, red]);

Note that the blue curve is between the two step functions colored yellow and red. If curve encloses infinite area then so does the larger step function because at all times it is greater than or equal to the curve. On the other hand, if the curve encloses a finite area, then so does the smaller step function. Simple enough. However, the upper step function is the lower one shifted one unit to the right. In other words, the upper one is the same as the lower one plus one additional term of a1 = 1. Consequently, the series converge or diverge as the integral does.

> Sum( f(k), k = 1..infinity): % = evalf(value(%));

> Int( f(x), x = 1..infinity): % = value(%);

> Sum( f(k+1), k = 1..infinity): % = evalf(value(%));

In this case, we can actually compute the integral and the sums. Note that all three converge, however they are not equal. The value of the integral is between the value of the sums from k = 1.. infinity and k = 2..infinity - just as we saw on the graph above. Thus the integral tells us about convergence & divergence but not exact values.

One important result that follows directly from the integral test, is that series of the form 1/k^p converge only for

p > 1, and diverges otherwise.

> assume( q > 0); p := q + 1; Int( 1/x^p, x = 1..infinity ): % = value(%);

> assume( q <= 0); p := q + 1; Int( 1/x^p, x = 1..infinity ): % = value(%);

Note that integral is finite for p > 1, and finite for p =< 1.

___________________________________________________________________________________

B. The Limit Comparison Test

___________________________________________________________________________________

Another important test is the Limit Comparison Test. A simplified version of this test asserts that if the ratio of the terms of two different sequences have a finite, non-zero limit, then both converge or both diverge. Lets examine why test works from a geometric point of view. Consider these three sequences.

bk is known to converge while ck is known to diverge. For now we are not sure if ak converges or diverges.

> a := n -> (3 + sqrt(n))/(n^2 - sqrt(n) + 5);

> b := n -> n^(-3/2);

> c := n -> 1/n;

Lets look at what is happening graphically. The red graph is subject ak, while bk is blue and ck is green.

> plot( [a(floor(x)), b(floor(x)), c(floor(x)) ], x = 2..5, color = [red, blue,green]);

From this graph it appears that ak is larger than both bk and ck. And since ck diverges, ak might diverge also!

> plot( [a(floor(x)), b(floor(x)), c(floor(x)) ], x = 5..20, color = [red, blue,green]);

That theory was premature. We now see that things are changing.

Ultimately, ak and bk are more or less in sync. We can see this we apply the Limit Comparison test and take the limits of the terms. This does prove the limit comparison test, but makes it plausible.

> limit( a(n)/b(n), n = infinity);

> limit( b(n)/c(n), n = infinity);

___________________________________________________________________________________

C. The Ratio Test

___________________________________________________________________________________

More generally, the ratio of consecutive terms is an expression. The ratio test requires us to take the limit of the absolute value of this ratio. When this limit is strictly less than 1, the series converges absolutely.

Another important test is the Ratio test. In this test, we take the limit of the absolute value of consecutive terms. If

> restart;

> Sum( a*r^n, n = 0..infinity): % = value(%);

> a := n -> (6/7)^n;

> a(n+1)/a(n);

> simplify(%);

> Sum( a(n), n = 0..infinity): % = value(%);

Here we create a generic geometric series, then look at an example for an = (6/7)^n. We take the ratio of consecutive terms, which is a constant of course. We then verify that the series converges.

> a := n -> (2^n) / n!;