Did you recognize the above relation from the drawings?If not, go back and take another look, or study my drawing.
Notice the three series that we have observed so far:
1/4 + (1/4)2 + (1/4)3 + . . . + (1/4)n + . . . = 1/3
1/3 + (1/3)2 + (1/3)3 + . . . + (1/3)n + . . . = 1/2
1/2 + (1/2)2 + (1/2)3 + . . . + (1/2)n + . . . = 1/1 = 1It would now be a good conjecture that given c is a natural number,
1/c + (1/c)2 + (1/c)3 + . . . + (1/c)n + . . . = 1/(c-1)There's only one thing wrong -- I don't know how to illustrate it using a visualization technique the way we did before. If you do, email me and I'll add it to the page. Until then, we'll have to use the traditional numerical methods and formulas.
Exploration 3Consider the series
1/5 + (1/5)2 + (1/5)3 + . . . + (1/5)n + . . .Enter the terms into a calculator summing after each entry. From our conjecture, we believe that the sums should come closer and closer to what number after each entry? Do they?
Use the square grid paper and design a drawing that will illustrate the sum of the infinite series above.
Let's use a very clever mathematical device to prove our conjecture. Let S be the sum of our series. Then S = 1/c + (1/c)2 + (1/c)3 + . . . Now here comes the clever part: multiply both sides of the equation by 1/c and line up like terms. Why do such a weird thing? Because it works!
S = 1/c + (1/c)2 + (1/c)3 + . . . (1/c)S = (1/c)2 + (1/c)3 + . . .
Now subtract the bottom equation to get
S - (1/c)S = 1/c
(1-1/c)S = 1/c
S = 1/(c-1), just as we suspected :-)
Now I wonder if the numerator has to be 1 or could it be any positive number (or even negative). Do keep in mind that |a/c| must be less than 1 for the series to have a finite sum. Why don't you try using our mathematical device above to see if
a/c + (a/c)2 + (a/c)3 + . . . + (a/c)n + . . . = a/(c-a)Now wouldn't that be too cool?
[To the Top] [Next - Links] [Back][More Fun Mathematics]
Copyright 1999-2008 Cynthia Lanius