Michael, a 5th grader, finds a Pattern in Infinite Series

After coloring squares for  -> 1 and -> 1/2 and 1/4+(1/4)^2+...-> 1/3, he came up with the generalization that 1/n + (1/n)^2 + (1/n)^3 + (1/n^4)… approaches the limit 1/(n-1).

Don showed Michael how to do this in Mathematica® and Derive. Michael  tried things to show how his rule works for big numbers..

The Greek symbol Sigma means sum. So this reads find the sum of 1/3 to the nth power as n goes from 1 to 2, to 3, ... to infinity and looks like this below

which approaches 1/2 as the limit.  Michael did the following in Derive.

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and  he realized that 1/n + (1/n)^2 + (1/n)^3 + (1/n^4)… approaches the limit 1/(n-1).

He even tried

  which converges to 0

Don has found that many of his students try to see what happens when they push the envelope!

A week or two later, Don asked Michael to see what happens if we start with 2/5.

In Derive  Michael did a number of infinite sums, starting with

2/5 , which went to 2/3 and

74/75,  which went to 74 and

7/676767677665883, which went to 7/676767677665876 and others,

and he figured out that

He also found, by himself, that the infinite sum, starting with 89/6 gets bigger and bigger, it diverges. and starting with 3/3, the series would also diverge, does not approach a number. So they talked about how A and B were related.

 Michael said for the infinite geometric series to converge, 

A must be < B .

In Derive we looked at A/B + (A/B)^2 + (A/B)^3 + … using the input below and got the answer on the right:    

Don never saw this, and didn't know what SIGN(x) meant, and Michael didn't know what LN meant, so it was interesting to figure out what this meant with Michael. First, the second fraction in the answer

 which is what Michael got above.

From the HELP section in Derive: "SIGN(x) simplifies to the sign of x.  Hence if x is a positive number, SIGN(x) simplifies to 1.  If x is a negative number, SIGN(x) simplifies to -1.  Since the sign of 0 is indeterminate, SIGN(0) simplifies to plus-or-minus 1". 

Now, if A<B, or A/B<1, then the LN (A/B) is a negative number (Don will work on logarithms with Michael soon), then SIGN(x) simplifies to -1. Then the top of the left fraction becomes B* e^ (-∞) = B*1/(e^∞) and the numerator goes to 0, and the answer becomes

Now, if A>B, or A/B>1, then the LN (A/B) is a positive number and SIGN(x) simplifies to 1. Then the top of the left fraction becomes B* e^ (∞) and the numerator goes to infinity, and the answer goes to infinity and the series is divergent.

Now, if A=B, or A/B=1, then the LN (A/B) is 0 and SIGN(x) is indeterminate and simplifies to ±1. Then the top of the left fraction becomes B* e^ (±∞) which doesn't mean anything to Don. He and Michael agreed that if A/B =1, the series is divergent.

Derive tries to answer all the conditions for A and B.

In contrast, Mathematica assumes A<B, because, it gives the output that is equivalent to what Michael got above :

Input:  Sum[(A/B)^n,{n,∞}]

Output:

Fine work Michael!!!