### The order of addition matters!

Now we are ready for our demonstration that the infinite sum:

1 - 1/2 + 1/3 - 1/4 + 1/5....... - 1/(2n) + 1/(2n +1)......

can be rearranged to equal any number that we chose. In other words, the order of addition matters!

The idea is this: choose any number in the real number system that you please. For example, choose, say "2".

Now since we picked a positive number, start with the positive terms.

Start with 1 + 1/3 + 1/5 + 1/7 and keep adding until you first get a number greater than 2.

How do you know that you can do this? Because 1 + 1/3 + ....+ 1/(2k+1) + 1/(2k+3) +....

is an infinite sum by the integral test. Exercise: find how many terms it takes to get bigger than 2.

Next start with the negative terms -1/2 - 1/4 - 1/6....and keep going until you first get a number less than 2. Why can this always be done?

Next start with the first positive term you didn't use in your first step and work your way past 2 again, and then use the negative terms to get less than 2 again, and keep repeating this process.

To see why this process yields a sum of 2, let s1, s2, s3,...denote the following partial sums:

s1 = sum of the first set of positives, s2 = sum of the first set of negative terms, s3 = sum of the second set of positive terms, and so on.

Notice that s1 can exceed 2 by at most some 1/n1. Then when we use s2, note that we only go at most 1/m1 past 2 in the negative direction. Then using s3, we exceed 2 by at most some 1/n2, where 1/n2 < 1/n1 (why does this inequality hold?) and then using s4, we go at most 1/m2 below 2 where 1/m2 < 1/m1.

So, given any interval about our chosen number 2, the partial sums must eventually lie within this interval (this is an "epsilon-delta" argument).

In other words, we have shown that our series has been rearragned to have partial sums which go "back and forth" about 2 and get arbitrarily close to 2.

1 - 1/2 + 1/3 - 1/4 + 1/5....... - 1/(2n) + 1/(2n +1)......

can be rearranged to equal any number that we chose. In other words, the order of addition matters!

The idea is this: choose any number in the real number system that you please. For example, choose, say "2".

Now since we picked a positive number, start with the positive terms.

Start with 1 + 1/3 + 1/5 + 1/7 and keep adding until you first get a number greater than 2.

How do you know that you can do this? Because 1 + 1/3 + ....+ 1/(2k+1) + 1/(2k+3) +....

is an infinite sum by the integral test. Exercise: find how many terms it takes to get bigger than 2.

Next start with the negative terms -1/2 - 1/4 - 1/6....and keep going until you first get a number less than 2. Why can this always be done?

Next start with the first positive term you didn't use in your first step and work your way past 2 again, and then use the negative terms to get less than 2 again, and keep repeating this process.

To see why this process yields a sum of 2, let s1, s2, s3,...denote the following partial sums:

s1 = sum of the first set of positives, s2 = sum of the first set of negative terms, s3 = sum of the second set of positive terms, and so on.

Notice that s1 can exceed 2 by at most some 1/n1. Then when we use s2, note that we only go at most 1/m1 past 2 in the negative direction. Then using s3, we exceed 2 by at most some 1/n2, where 1/n2 < 1/n1 (why does this inequality hold?) and then using s4, we go at most 1/m2 below 2 where 1/m2 < 1/m1.

So, given any interval about our chosen number 2, the partial sums must eventually lie within this interval (this is an "epsilon-delta" argument).

In other words, we have shown that our series has been rearragned to have partial sums which go "back and forth" about 2 and get arbitrarily close to 2.

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