Limits as x --> Infinity

First week of calculus, and the talk is all about limits and infinity...

They say that one of Leibniz's earliest explorations into limits and infinite series was in finding the sum of the reciprocals of the triangular numbers, or $\sum_{1}^{\infty}\frac{2}{(n)(n+1)}\$



The triangular numbers are the numbers that can be made by adding the counting integers

1, 1+2=3, 1+2+3=6, etc.. they also appear as the third number in on each row of Pascal's arithmetic triangle, so they are the combination of n things taken two at a time. The n^th triangular number then, is $\dbinom{n}{2}$ = $\frac{(n)(n+1)}{2}$



The sequence of reciprocals would then $\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{(n)(n+1)}$



But if we write them out using the $\frac{2}{(n)(n+1)}$= $\frac{2}{n}}$ - $\frac{2}{n+1}$ model for each term, then something interesting happens...

writing 1 as $\frac{2}{1}-\frac{2}{2}$ may seem silly at first, but if you follow that with 1/3 = $\frac{2}{2}-\frac{2}{3}$ you realize that when you add these together, the alternating terms eliminate each other and the sum of the first two terms is just 2/1 - 2/3 or 1 ¹/₃. Ok, the worlds slowest way to add one to one-third.. but now lets add the next few values 2/1-2/2 + 2/2-2/3+2/3-2/4+2/4-2/5 + 2/5 - 2/6 we see that all the underlined stuff disappears and we are left with the first term, 2, minus the last term, which is 2/(n+1) for whatever term number we are at.. so the sum of the first five terms, 1+1/3+1/6+1/10+1/15 = 2-2/6 or 1 ²/₃.

and if we wanted the first 20 terms we could just do 2 - 2/21 or 1 ¹⁹/₂₁. and if we added them up forever???? well what happens to ²/_(n+1) as n gets bigger and bigger??? ....so the infinite sum will just approach a limit of two.



After I did this in class recently, I found this nice graphic demonstration at the wolfram demonstrations page;...a proof without words.





"This problem was historically significant as it served as in inspiration for Leibniz to explore many more infinite series. Since he successfully solved this problem, he concluded that a sum could be found of almost any infinite series. (Boyer, 446-447)" ( Worcester Polytechnic Institute )

Triangular Numbers

A friend reminded me that I hadn't explained triangular numbers here or on my web site, so taking care of both in a limited way today...

The triangular numbers are known back to before Pythagoras (500ish BCE). They are simply the sums of the natural or counting numbers 1 + 2 + ... + n is the n^th triangular number, so the sequence of triangular numbers is 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, etc. An explicit formula for the triangular numbers is T_n = n(n+1)/2. For example, T₁₀= (10)(11)/2 = 55.

The triangular numbers appear as the third number in each row of Pascal's triangle.

This means that they are also expressible as (n+1 choose 2)

One of the interesting properties of the triangular numbers is that any two consecutive ones form a square number.... in the image below, T₄ (in red) is added to T₅ in blue, to form S₅, the fifth square number or 25.

It is also known that any even perfect number, and all the ones we know of are even, is a triangular number. The first few are 6, 28, 496.

It is also known that the digit root in base ten of any triangular number is either 1 or 0. That means that all triangular numbers are either divisible by three, or have a remainder of one when divided by three. An easy test of any number to see if it is triangular is to multiply by eight and add one... if the result is a perfect square, then the number is triangular.

It is also interesting that the square of a triangular number T_n is equal to the sum of the cubes of the natural numbers from 1 to n,

....

And one final note on the triangular numbers, the infinite sum of their reciprocals is an integer, ... how sweet, and unexpected, is that? I think you can figure out what it is..but if you get stuck and need a hint, or the answer... just drop a note

Pick Two

A colleague from Colorado sent me an interesting probability problem the other day. I like it because it illustrates one of those serendipitous qualities of mathematics. Here is the problem. A Jar has a mixture of Red and White balls so that if you withdraw two, the probability of getting two alike, or two of different color are both equal to one-half. You may want to stop and try it before you read on.

Ok, so we let r = the number of red ones and W be the number of white ones, and the total is r+w. So how could we draw one of each color? Well, red first, and then white, or white first and then red. If we find the probability of each of these conditional events and add them up, that will have to equal 1/2. Ok, the probability of red on the first draw is r/(r+w), and on the second ball the probability is w/(r+w-1) since one of the red balls will be missing. The opposite order is exactly the same with the w and r reversed, so the probability of getting one of each color is 2rw/[(r+w)(r+w-1)]. Setting that equal to 1/2 we get

If we exapand (r+w)² and subtract the 4rw we get 0=r²-2rw +w²-r-w. NOTICE the symmetry, we could exchange r and w and get the same equation. We know right off that any solution (a,b) will have another soltuion (b,a).

One of the things that is often hard for students is to think of one variable as a constant and the other as a variable. I like to use the word "pronumeral", like a pronoun only instead of him or her we say "that number". It is like a variable that doesn't vary, we just don't know what it is in a particular case. So think of w as if it were fixed. We have that many white balls in the jar and we are wondering how many red can be put in to make the problem work... see it.. w is a "fixed" unknown, but r is going to "vary". That makes the equation a quadratic in r; Ar² +Br+c=0 where A=1, B= -2w-1, and C=w²-w.

We can solve this using the quadratic formula, but if this solution is going to be a rational number, and the number of balls in a jar must be rational, then the discriminant, the expression under the squre root radical in the quadratic forula, B²-4AC, must be a perfect square. B²= 4w²+4w+1 and 4AC= 4w²-4w; so B²-4AC= 8W+1. If there is a rational solution, it must be when 8W+1 is a perfect square. Wait, I know this one! That's a problem from number theory. The numbers that make 8W+1 a perfect square are called triangular numbers; 1, 3, 6, 10, 15. They are the sum of the first n counting numbers. But a neat thing happens if we plug 1 in for W, the solution for r is 3.... and if we use 3 for w, r=6. Each time we substitute one of the triangular numbers into the quadratic, the next comes out as a solution. So the probability of drawing two balls of the same color, (or of two that are not alike) will equal 1/2 whenever the number of balls of each color are consecutrive triangular numbers. A very geometric solution to a very algebraic question.