Grandi’s series is divergent. With divergent series, and also conditionally convergent series, you get different “answers” by grouping the terms in different ways.

I agree, Grandi's Series is divergent. And I agree that you get different answers by grouping in different ways.

But please notice I didn't "group the terms" in Grandi's Series as you say. What I did was I lined the terms up for two infinite series (one was Grandi's series, the other was a slightly modified version of the same series). The two series are the same length because they have a simple 1 - 1 correspondence which lasts all the way to infinity.

I think it is quite possible to get different answers like you stated:

- if you group terms or

- if you don't have a 1-1 correspondence (i.e. you skip a term, or clump a a couple of terms or more, or... something) or

- if you stop prematurely (i.e. you convert the infinite series to a finite series),

but I didn't do any of those things.

Could you show me a different answer for Grandi's Series other than 1/2? Of course, it would be best if you used the same technique I used i.e. lining up two infinite series using a demonstrable 1-1 correspondence. It would really help me to understand!

thanks,

John

PS I really don't know if my "proof" is right. I'm not insisting or anything like that.

It's just that I haven't broken any axioms or operational rules in Integer Algebra. As far as I know, each operation I'm using has no flaw. So, to me, the "proof" looks good.

PSS I do realize I am making these assumptions:

1) I assume that it is ok to sum a1 + a2 + a3... to infinity. As someone pointed out, "+" is defined as a binary operation, so it can only take two operands. So a1 + a2 + a3 ... is really a1 + (a2 + (a3 + .... ))))) <-- an infinite number of parentheses here -- which may cause a problem. I don't see it, but maybe there is one.

2) it's ok to sum the terms of two infinite series if they are the same length, and there's no shifting or misalignment issues way down the line, i.e. the two series start aligned and stay aligned to infinity.

3) the sum of 0+0+0... to infinity is 0. Again it seems reasonable, but see #1

4) algebraic manipulations like "2S = 1 --> divide both sides by 2 --> S = 1/2." are ok to do when S started off as an infinite series. I know these manipulations are ok to do if S is a simple, finite number. But it is possible, I guess, that if S actually has the value "infinity" then you do an operation like "2S=1 --> divide both sides by 2", it will result/report as a simple, finite number. It seems very, very unlikely that this would be the case, but I don't know for sure.

5) others?

If you guys see a flaw in any step or operation I've done or any other assumptions, please let me know!

]]>Write for the equation.

One simple formula.

If you use the solutions of the equation Pell. Where

ask yourself.Make the change.

Then decisions can be recorded.

]]>PiperThesis

http://www.theliberatedmathematician.com/wp-content/uploads/2015/11/PiperThesisPostPrint.pdf

]]>When you draw a hexagon, draw three lines from the center in even spaces./ Draw a cube in 3-D and you will see a hexagon on the outside.

7&7=?

= parallelogram or rhombus ( turn one of the sevens' upside-down!)

Can you post even more tricks?

Images would teach others. TY

]]>So fractions yes. Negatives probably not

Bob

]]>```
1: 0
2: 0 0
3: 0 1 0
4: 0 1 0 0
...
```

1 is divisible by 1.

2 is divisible by 1 and 2.

3 is divisible by 1 and 3, and has a remainder of 1 when divided by 2.

4 is divisible by 1, 2 and 4, and has a remainder of 1 when divided by 3.

..

Notice that the modulos are in the reverse order, such that the first number on a line is n modulo n, and the next is n modulo (n-1).

This is a screenshot of a text editor with the pattern for the first 300 numbers or so, with 0 highlighted.

You get a bunch of lines! The leftmost line is the n%n line. All numbers are divisible by themselves. The next line is n%(n/2), all numbers that are divisible by 2, are also divisible by themselves divided by two. The pattern continues like this, the next line being n%(n/3).

```
..
18: 000 001 002 003 004 005 006 007 008 000 002 004 000 003 002 000 000 000
19: 000 001 002 003 004 005 006 007 008 009 001 003 005 001 004 003 001 001 000
20: 000 001 002 003 004 005 006 007 008 009 000 002 004 006 002 000 000 002 000 000
21: 000 001 002 003 004 005 006 007 008 009 010 001 003 005 000 003 001 001 000 001 000
..
```

Notice that between the 1 and 2 line, the modulos are incrementing by one up to ceiling(n/2-1). After the 2 line, everything increments by 2.

You also see a example of diagonal incrementation lines, that is, you can pick any of the modulo numbers, go one down and one to the right, and that number is going to be one greater than the number that you came from, unless it is zero.

Consider 6:

```
06: 000 001 002 000 000 000
07: 000
08: 000
09: 000
10: 002
11: 001
12: 000
13: 006
14: 006
15: 006
...
```

If you flip this sequence 90 degrees, it is going to replicate:

6 is divisible by 1, therefor 7 is divisible by 1.

6 is divisible by 2, therefor 8 is divisible by 2.

6 is divisible by 3, therefor 9 is divisible by 3.

6 has a remainder of 2 when divided by 4, therefor 10 has a remainder of 2 when divided by 4.

6 has a remainder of 1 when divided by 5, therefor 11 has a remainder of 1 when divided by 5.

6 is divisible by 6, therefor 12 is divisible by 6.

after this point, 6 is just going to repeat infinitely downwards. This pattern is true for all numbers, and primes are just the unlucky numbers that didn't get any zeroes.

I am sure that there are a lot more patterns that I didn't cover, so I'm gonna leave that to the comment section

Haskell program that generates the first 1000 lines in case anyone want it:

```
mods x = map (mod x) [1..x]
mods' = map mods [1..]
show' x = replicate 3 - length (show x) '0' ++ show x
main = putStrLn . unlines . map (unwords.map show'.reverse) $ take 1000 mods'
```

1. There are regular gaps of **4** between primes

2. These gaps of **4** have varying remainders for 4, (either 1 or 3)

A= 3x5x7x11x13x17............(x next prime in series)

x = the next prime above highest prime in A used

p and p+4 are prime

(A - p)/2 = twin prime

(A-(p+4))/2 = twin prime

**Example:**

A= 3x5x7= 105

x=11

A has remainder=1 for 4

2 Primes = 67 and 71

67 and 71 have remainder=3 for 4, which is different from A

(105-67)/2 = 19

(105-71)/2 = 17

Twin primes!

A= 3, 15, 105, 1155, 15015 x next prime in the series................

p= prime or not factorable by any primes in A

and a gap of 2, therefore a p+1 can NOT be a prime because p+2 can not be prime

and a gap of 4, therefore a gap of 4 can not occur because p+2 can not be prime

OR and a gap of 4, therefore a gap of 4 can not occur because p+1 can not be prime

With a gap of 4, one number minused from A will be divisible by 4 or 2. If one is divisible by a higher than 4, the other can't be because the gap would have to be higher than 4 for the remainders to match. Therefore we can express a gap of 4 using the above equations, where if one number is divisible by a greater than 4 this will be p+1 an even no. and p is what was divisible by 4 but no higher.

and a gap of 8, therefore a gap of 8 can not occur because p+4 can not be prime

and a gap of 8, therefore a gap of 8 can not occur because p+2 can not be prime

and a gap of 8, therefore a gap of 8 can not occur because p+1 can not be prime

With a gap of 8, one number minused from A will be divisible by 2,4 or 8. If one is divisible by a higher than 8, the other can't be because the gap would have to be higher than 8 for the remainders to match. Therefore we can express a gap of 8 using the above equations, where if one number is divisible by a greater than 8 this will be p+1 an even no. and p is what is divisible by 8 but no higher.

Divide A by 3 and multiply by 2

another p

and a gap of 6, therefore a gap of 6 can not occur because p+2 can not be prime

With a gap of 6, one number minused from A will be divisible by 3. If one is divisible by a higher than 3, the other can't be because the gap would have to be higher than 6 for the remainders to match. Therefore we can express a gap of 6 using the above equation, where if one number is divisible by a greater than 3 this will be p+2 and p is what is divisible by 3 but no higher.

Divide A by 5 and multiply by 2

another p

and a gap of 10, therefore a gap of 10 can not occur because p+2 can not be prime

With a gap of 10, one number minused from A will be divisible by 5. If one is divisible by a higher than 5, the other can't be because the gap would have to be higher than 10 for the remainders to match. Therefore we can express a gap of 10 using the above equations, where if one number is divisible by a greater than 5 this will be p+2 and p is what is divisible by 5 but no higher.

**..........And so on until a gap that must be >2 becomes a gap that must be >infinity, there therefore must be an infinite No. of twin primes.**

]]>

check p is not a square

where a+b is a factor and a-b is another factor

if p is prime you will not find any factors

if p is composite you will find factors.

Example:

p=33

27*33=891

46+35=81=27*3 46-35=11

factors for 33 are 3 and 11]]>