by Soo Jin Lee and Jaehong Shin
For last three days, we have investigated infinite series with some examples.
In addition to that we came upa good conjecture that given c is a natural number,
(1/c) + (1/c)^2 + (1/c)^3 +... + (1/c)^n +... = 1/(c-1)
First, let us try one more investigation.
1. Consider the series
(1/5) + (1/5)^2 + (1/5)^3 + ... + (1/5)^n +...
Enter the term 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?
2. Use the square grid paper and design a drawing that will illustrate the sum of the infinite series above.
3. 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 +... + (1/c)^n +... Now here comes the clever part: multiply both sides of the equation by 1/c and line up like terms. Why do such a wierd thing? Because it works.
i.e. 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.
4. 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).
I will leave this as an exercise.