北邮高数授课讲义

A Bouncing Ball

Lecture 1 Series of Constant Terms

Drop a ball from H meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than 1. find the total distance the ball travels up and down. It is easy to see that the total distance is s= H+ 2 Hr+ 2 Hr 2+ 2 Hr 3+

The question is that how to calculate the q sum with infinite terms. It is well known that, if there has finite terms sn= H+ 2 Hr+ 2 Hr 2+ 2 Hr 3+ 2 Hr (1 r n )=H+ 1 r+ 2 Hr n

It is clear that lim sn= H+ n→∞

2 Hr (1 lim r n ) n→∞

1 r

=H+

2 Hr . 1 r 2

1

Repeating Decimals Express the repeating decimal 5.23 23 23… as the ratio of two integers. Solution 5.23 23 23...= 5+ = 5+

Infinite Series Definition The following representation a1+ a2++ an+

or

∑a n=1

∞

n

23 23 23+++ 100 (100)2 (100)3 23 1 1+ 1++ 100 100 100 2

which consists of all the terms of a sequence{an} and connected successively by the plus sign, is called an infinite series or simply a series, and an is called the general term of the series.

If the terms of the sequence are numbers then the series is called a q series of constant terms. If there is little chance of confusion, it may be called simply a series. It is convenient to use sigma notation to write the series as A useful shorthand when summation from 1 to∞ is understood

= 5+

23 1 23 518= . = 5+ 100 0.99 99 99

∑a n=1

∞

n

,

∑a k=1

∞

k

,

or

∑a

n

3

4

Infinite Series A finite sum of real numbers always produces a real number, but an infinite sum of real numbers is something else entirely. This is why we need a careful definition of infinite series. We begin by asking how to assign meaning to an expression like 1+ 1 1 1 1++++ 2 4 8 16

Infinite Series ∑a n=1 ∞

n

=1+

1 1 1 1++++ 2 4 8 16

Partial Sum First: Second: Third: …

Value

s1= 1 1 s2= 1+ 2

21 2 2 1 2 1 4

The way to do so is not to try to add all the terms at once (we cannot) but rather to add the terms one at a time from the beginning and look for a pattern in how these"partial sum" grow.

s3= 1+

1 1+ 2 4

…

nth

sn= 1+

… 1 1++ 2 4+ 1 2 n 1

2

1 2n1

5

6