Module 10 : Serious About Series

1001 : Series & Convergence

O B J E C T I V E

In this module, we will examine sequences and series from both a numerical and geometric point of view. We will look at convergence of sequence and series, and several convergence tests for series.

S E T U P

In this project we will use the following command packages. Type and execute this line before beginning 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;

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A. Sequences Numeric & Geometric

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In a sense, a sequence is the range of a function defined only on the positive whole numbers. A function can also be thought of as a numbers. Here we define the function a which is only defined for non-negative values of k. There are two ways to write ouit the sequence using the $ operator and the seq command.

defining a sequence as a function

> a := k -> ((-1)^k) * (2^k - 1)/(k! + 2);

a := proc (k) options operator, arrow; (-1)^k*(2^k-...

listing the first 12 elements

> a(k) $ k = 1..12;

-1/3, 3/4, -7/8, 15/26, -31/122, 63/722, -127/5042,...

...in two different ways

> seq( a(k), k = 1..12);

-1/3, 3/4, -7/8, 15/26, -31/122, 63/722, -127/5042,...

expressing the numbers in decimal format

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

-.3333333333, .7500000000, -.8750000000, .576923076...
-.3333333333, .7500000000, -.8750000000, .576923076...

We can also see what the limit of the sequence is, if the limit exists, and plot the sequence graphically.

> Limit( a(k), k = infinity): % = value(%);

Limit((-1)^k*(2^k-1)/(k!+2),k = infinity) = 0

> plot( a(floor(x)), x = 1..30 );

[Maple Plot]

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B. Series & Convergence

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If we add the terms in a sequence we get a series.

FINITE & INFINITE

We can add up a finite or infinite number of terms in a sequence to get a finite or infinite series.

A sum can contain a finite or infinite number of terms. Even when the sum is adding an infinite number of terms, the result can be finite.

> Sum( 3*k^4 - 5*k, k = 1..30): % = value(%);

Sum(3*k^4-5*k,k = 1 .. 30) = 15819672

> Sum( k/10000, k = 1..infinity): % = value(%);

Sum(1/10000*k,k = 1 .. infinity) = infinity

> Sum( 1/k^6, k = 1..infinity): % = value(%);

Sum(1/(k^6),k = 1 .. infinity) = 1/945*Pi^6

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

Sum(1/k,k = 1 .. infinity) = infinity

SEQUENCE OF PARTIAL SUMS

Every infinite series can be thought of as a an infinite sequence. Let S1 = a1, S2 = a1 + a2, S3 = a1 + a2 +a3 ,...etc. Each Sn is called a "partial sum" because it made up of only the sum of the first n terms. The limit of the sequence of partial sums {Sn} is the same as the infinite series.

> a := k -> 1/k^(3/2);

a := proc (k) options operator, arrow; 1/(k^(3/2)) ...

> S := n -> sum( a(k), k = 1..n);

S := proc (n) options operator, arrow; sum(a(k),k =...

> array([seq( [ 5*k, evalf( a(5*k) ), evalf( sum( a(j), j = 1..(5*k))) ], k =1..12)]);

matrix([[5, .8944271912e-1, 1.760446199], [10, .316...

The first column is k which increases in increments of 5. The second column is ak, and the third is Sk. Note that the ak's are approaching 0, while the Sk appear to approach some son-zero limit.

To help make the difference between a sequence and its corresponding series more apparent , lets look at both graphically.

> plot( { a(floor(x)), sum( a(k), k = 1..floor(x) ) }, x = 1..30);

[Maple Plot]

The terms of the sequence are decreasing to zero while the partial sums converge to some number.

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C. Slow Divergence

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Each when a series diverges to infinity, the journey may be quite leisurely. Lets examine the harmonic series, which is the sum of the reciprocals of the natural numbers. First we'll add up the first 100 reciprocals and see how close we are to infinity.

> Sum( 1/k, k = 1..100); value(%);

Sum(1/k,k = 1 .. 100)

14466636279520351160221518043104131447711/278881500...

The evalf command forces a decimal answer.

> evalf(%);

5.187377518

Not too close! Let add up the first 1,000 terms and see what progress is made.

> Sum( 1/k, k = 1..1000); % = evalf( value(%));

Sum(1/k,k = 1 .. 1000)

Sum(1/k,k = 1 .. 1000) = 7.485470861

disappointingly small! Surely by the time we add a million terms, we must be getting significantly large!

> Sum( 1/k, k = 1..1000000); % = evalf( value(%));

Sum(1/k,k = 1 .. 1000000)

Sum(1/k,k = 1 .. 1000000) = 14.39272672

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