## Good Mathematics for computing to infinity

Author:
Published: 2013-03-27
Words: 4465

# An Astounding Error: 1+2+3+4+5+6+... equals -1/12, really?

aka. Good Mathematics for computing to infinity

2013-03-20

Today’s Mathematics seem to say that the sum of all the positive integers equals:

1+2+3+4+5+6+... = -1/12

cf. "ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = -1/12" https://www.youtube.com/watch?v=w-I6XTVZXww

cf. "Why -1/12 is a gold nugget" https://www.youtube.com/watch?v=0Oazb7IWzbA

So, what is an infinite sum of integers, turns to be in this result:

1. a fraction,

2. a negative value,

3. a finite value

But, by taking any 1 of those 3 points, we can assume that writing this equality is an error, and that it must be a false statement.

So, what happened? Let’s follow the given demonstration:

S0 = 1 -1 +1 -1 +1 -1 +...

S0+S0 = 1 -1+1 +1-1 -1+1 +1-1 -1+1 +...

2.S0 = 1 + 0

S0 = 1/2

The trick on line 2 is to shift on the right all the sum, to obtain zero on that part. It's a very clever trick and there's no mistake to use this idea.

S1 = 1 -2 +3 -4 +5 -6 +...

S1+S1 = 1 -2+1 +3-2 -4+3 +5-4 -6+5 +...

S1+S1 = 1 -1 +1 -1 +1 -1 +...

2.S1 = S0

S1 = 1/4

On line 2 the same trick is used: a shift on the right of all the sum. And again, it's very neat.

S2 = 1 +2 +3 +4 +5 +6 +...

4.S2 = 4 +8 +12 +...

S2-4.S2 = 1 +2-4 +3 +4-8 +5 +6-12 +...

S2-4.S2 = 1 -2 +3 -4 +5 -6 +...

-3.S2 = S1

-3.S2 = 1/4

S2 = -1/12

On line 3 subtracting 4.S2 is a beautiful idea, and it looks pretty good. But the problem of the fractional result when summing only integers appears in every result, so they must all be false.

So where are the errors? Line 1 is a statement so it can’t be false. Line 2 should contain the first error, because lines 3 and 4 look legitimate, normal mathematics, without involving infinity. Now, about line 2:

1. on the left side, we have "1", which is ok,

2. in the middle we have a bunch of "-1+1 +1-1", which is also ok,

3. on the right side, we have "..." but, what does "contain" those characters? What do they hide? Let’s write them down, to fully understand what we are doing here

Infinity, is not a number: it’s more an idea. We can think about doing something "to infinity", it’s a mental construct, it’s not real (cf. Zeno's paradoxes - Achilles and the Tortoise http://en.wikipedia.org/wiki/Zeno's_paradoxes#Achilles_and_the_tortoise ). So we can write for example "sum(something(x)) for x from 1 to infinity" but the result when we add something "to infinity" won't converge to a value if the term does not converge to zero, and it's even not sufficient. This prevents from obtaining a finite result for S1 and S2. We will transform the sums from 1 to "n" into functions depending on "n", and because the result will be true for any "n" (positive non null integer), it will be true for infinity as well ("∞").

So, let’s write down this "infinite" term and see what it hides:

• let odd() be a function that return 1 if the value taken is odd, and 0 else.

• let even() be a function that return 1 if the value taken is even, and 0 else.

• let "n" be any positive non null integer

fodd-even(x) = odd(x) - even(x)

fodd-even(x) = -1.(-1)^x

S0(n) = sum(1.fodd-even(x)) for x from 1 to n

S0(n)+S0(n) = 1 + 0 + 1.fodd-even(n)

S0(n) = 1/2 + 1/2.fodd-even(n)

S0(∞) = 1/2 + 1/2.fodd-even(∞)

We do the same trick on line 2 as the error demonstration: we shift to the right the second S0, so that every term equals zero in between (n-1 terms), and we free the first item of the list "1" and also we free the "last" item of the infinite sum "1.fodd-even(∞)". That last part was missing in the previous demonstration, and that’s where the error comes from. The error demonstration is only returning the finite part of the sum. But removing the "infinite part" of the sum (the part "to infinity") is nonsense, even if it's small in S0.

So, as we see, the result is not a fraction, even if it contains fractions. And it’s not a number: it’s a function that gives integers: either 0 or 1 if the value "taken to infinity" is even or odd, respectively.

Comparing with the previous error result, when the term "1/2.fodd-even(n)" was forgotten, the size of the error was "+- 1/2".

What’s next is slightly more complicated:

S1(n) = sum(x.fodd-even(x)) for x from 1 to n

S1(n)+S1(n) = 1 - [sum(1.fodd-even(x)) for x from 1 to (n-1)] + n.fodd-even(n)

S1(n)+S1(n) = 1 - (S0 - 1.fodd-even(n)) + n.fodd-even(n)

2.S1(n) = 1 - (1/2 - 1/2.fodd-even(n)) + n.fodd-even(n)

S1(n) = 1/4 + (1/4 + n/2).fodd-even(n)

S1(∞) = 1/4 + (1/4 + ∞/2).fodd-even(∞)

We do the same trick on line 2 as the previous demonstration: we shift to the right the second S1

From line 2 to line 3, we use this:

[sum(1.fodd-even(x)) for x from 1 to (n-1)] + 1.fodd-even(n) = S0

[sum(1.fodd-even(x)) for x from 1 to (n-1)] = S0 - 1.fodd-even(n)

Comparing with the previous error result, when the term "(1/4 + n/2).fodd-even(n)" was forgotten, the size of the error "to infinity" is about "+- ∞/2" and this time, it’s quite a big error. And again, the error result found the fix part of the sum, not the infinite one.

This result is working for any "n" (positive non null integer), that's why it's still true to infinity.

Now with S2:

2013-03-21

S2(n) = sum(x) for x from 1 to n

S2(2.n) = sum(x) for x from 1 to 2.n

S2(2.n) = [sum(x) for x from 1 to n] + [sum(x) for x from n+1 to 2.n]

S2(2.n) = [sum(x) for x from 1 to n] + [sum(n+x) for x from 1 to n]

S2(2.n) = [sum(x) for x from 1 to n] + [sum(x) for x from 1 to n] + [n.sum(1) for x from 1 to n]

S2(2.n) = 2.S2(n) + n²

S2(2.∞) = 2.S2(∞) + ∞²

S2(2.n) = [sum(2.x) for x from 1 to n] + [sum(2.x - 1) for x from 1 to n]

S2(2.n) = [sum(2.x) for x from 1 to n] + [sum(2.x) for x from 1 to n] - [sum(1) for x from 1 to n]

S2(2.n) = 2.[sum(2.x) for x from 1 to n] - n

S2(2.n) = 4.S2(n) - n

S2(2.∞) = 4.S2(∞) - ∞

4.S2(n) - n = 2.S2(n) + n²

2.S2(n) = n² + n

S2(n) = 1/2.n² + 1/2.n

S2(∞) = 1/2.∞² + 1/2.∞

This idea of the demonstration is to compute the sum by 2 different manners, we don’t even need S0 or S1. On the first part of the demonstration, we slice the sum by the middle and translate its right part to make the left part appear. On the second part of the demonstration, we create the same sum by adding all the odd numbers and all the even numbers, and again we make the left part appear in the right part (the sum of all odd numbers contains the sum of all even numbers, plus something...).

We could have computed directly S2 to n with the same ideas, but the problem when we write n/2 is that the slicing method depends on the fact that n is even or odd (and that's why we could have use S1), so we end up with 2 different results that compute the values one for n even and one for n odd. So rather than that, we compute S2 to 2.n, because what's true to n is still be true to 2.n, obviously.

About the result, again it contains fractions but they don’t display in the ending values. Those values can be checked for any "n" (positive non null integer), they are the good ones: S2(n) = 1/2.n² + 1/2.n

The size of the error "to infinity" is now around "1/2.∞² + 1/2.∞" and this time, it’s very very big.

A graphical representation also gives a demonstration of the result:

2013-03-26

2013-03-21

About the error result, "-1/12", well, we don’t see it anywhere in the result, there is no "fix part" in the sum, the curve starts at 0. "-1/12" is probably an artefact: a succeeding method to compute the fix part of the sum would have found "0". Let’s display the result:

If we derivate it, we obtain:

https://www.wolframalpha.com/input/?i=derivative%281%2F2.x*x+%2B+1%2F2.x%29

dS2(n) / dn = n + 0.5

So the minimum of the curve is for n = -1/2, and:

S2(-1/2) = 1/2.(1/2)² - 1/2.1/2

S2(-1/2) = -1/4

There is still no "-1/12".

In the search of "-1/12"

2013-03-24

Let’s use the beautiful idea of "S2(n) - 4.S2(n)" to see what we can find:

S2(n) - 4.S2(n) = [sum(x) for x from 1 to n] - 4.[sum(x) for x from 1 to n]

S2(2.n) - 4.S2(2.n) = [sum(x) for x from 1 to 2.n] - 4.([sum(x) for x from 1 to n] + [sum(x) for x from n+1 to 2.n])

-3.S2(2.n) = S1(2.n) - 4.[sum(x) for x from n+1 to 2.n]

-3.S2(2.n) = S1(2.n) - 4.[sum(n+x) for x from 1 to n]

-3.S2(2.n) = S1(2.n) - 4.(S2(n) + n²)

-3.(2.S2(n) + n²) = S1(2.n) - 4.S2(n) - 4.n²

-6.S2(n) - 3.n² = S1(2.n) - 4.S2(n) - 4.n²

-2.S2(n) = S1(2.n) - n²

S2(n) = -1/2.[1/4 + (1/4 + n).fodd-even(2.n)] + 1/2.n²

S2(n) = -1/2.[1/4 -1.(1/4 + n)] + 1/2.n²

S2(n) = 1/2.n² + 1/2.n

For the error result, the forgotten part (to 2.n) was:

- 4.[sum(x) for x from n+1 to 2.n]

And remarkably, we have:

fodd-even(2.n) = -1

There is still no "-1/12" anywhere. It was just an error. Euler was wrong on this.

2013-03-23

<< What is common in all those demonstrations is "∞". It's not a value, it's not even the idea of a value, "doing something to infinity" is the idea, it's a process. So "∞" is a variable that can take any integer value (just like a "x"), so we can change it to "y" or "n" or whatever and the results will still be all true. Then what will be true for any value, will be still true for infinity. >>

In fact, I first wrote this document with a "∞" everywhere (and no "n"), then I changed it to "n" after writing the previous paragraph. "n" and "∞" are equivalent here, because I respected the value of "∞" for itself: an unknown value, accepting to write things like "∞/2", "2.∞", "1/4 + ∞", or even this odd "fodd-even(∞)".

If we take the idea of infinity, we would have replaced "2.∞" with "∞", because nothing is bigger than infinity, they all "look as big". But writing this leads to quite huge mistakes, as we just demonstrated, so it has to be avoided.

Deeper on the error

2013-03-20

Let’s look at another serie: S11 = sum(1) for x from 1 to ∞, so that S11 = ∞

If we continue to forget the "right part" of the infinite sum, then we would be able to write:

If we shift 1 to the right:

S11-S11 = 1 + 0, so 0 = 1 ??

And if we shift ∞/2 to the right:

S11-S11 = ∞/2 + 0, so 0 = ∞/2 ??

Now, if we correctly write the right infinite part:

With 1 shift to the right:

S11-S11 = 1 + 0 - 1 = 0

With ∞/2 shifts to the right:

S11-S11 = ∞/2 + 0 - ∞/2 = 0

So, losing the right part of an infinite sum when we use a shift to the right "is zooming" on the left part, the first element of the serie, and brings it in the result, while forgetting the right part which is, obviously for S1 and S2, the greatest.

Conclusion

2013-03-23

What's demonstrated here is that the 3 values widely accepted as the results of S0, S1 and S2, are wrong. And where the error comes from is also demonstrated.

It doesn't add anything (as a result) to the Mathematics, it only removes a mistake. Maybe the method used here contains interesting aspects, I don't know about it.

But it just seems so straightforward, so logical. I had the idea the very first day I saw in school those infinite sums, in 1998.

Thomas Godart
written from the 2013-03-20
to the 2013-03-24

Whoops! The same mistake can be found in other results:

1+4+8+16+... equals -1, really?

2013-03-24

The given demonstration:

S3 = 1 + 2 + 4 + 8 +...

S3 = 1 + 2.(1 + 2 + 4 +...)

S3 = 1 + 2.S3

S3 = -1

The good result for "n" any positive non null integer:

S3(n) = sum(2^x) for x from 0 to n

S3(n) = 1 + 2.[sum(2^x) for x from 0 to n-1]

S3(n) = 1 + 2.([sum(2^x) for x from 0 to n] - 2^n)

S3(n) = 1 + 2.(S3(n) - 2^n)

S3(n) = -1 + 2.2^n

S3(∞) = -1 + 2.2^∞

So the size of the error "to infinity" is about "2.2^∞".

1 + x + x² + x³ +... equals 1 / (1 - x), really? (for |x| < 1)

2013-03-24

The error result:

S4 = 1 + x + x² + x³ +...

S4 - 1 = x.S4

S4 = 1 / (1 - x)

The good result for any non null "x":

S4(x,n) = sum(x^y) for y from 0 to n

S4(x,n) / x = [sum(x^y) for y from -1 to (n-1)]

S4(x,n) / x = 1/x + S4(x,n) - x^n

S4(x,n) . (1 - x) = 1 - x^(n+1)

S4(x,n) = [1 - x.x^n] / (1 - x)

S4(x,∞) = [1 - x.x^∞] / (1 - x)

Another method:

S4(x,n) = sum(x^y) for y from 0 to n

S4(x,n) . x = [sum(x^y) for y from 1 to (n+1)]

S4(x,n) . x = -1 + S4(x,n) + x^(n+1)

S4(x,n) . (x - 1) = -1 + x^(n+1)

S4(x,n) = [x.x^n - 1] / (x - 1)

S4(x,∞) = [x.x^∞ - 1] / (x - 1)

0.999... = 1, really ?

2013-03-24

The good result:

0.999... to n = 9.[S4(1/10, n) - 1]

0.999... to n = 9.[[1 - (1/10).(1/10)^n] / (1 - 1/10)] - 9

0.999... to n = 10.[1 - (1/10).(1/10)^n] - 9

0.999... to n = 1 - (1/10)^n

0.999... to ∞ = 1 - (1/10)^∞

It shouldn't be a surprise that 2 different numbers are proven to be different. One is an integer, and the other is not: it's a real number.

Trying to measure the distance between 2 infinitely close numbers leads to an infinitely small distance, obviously: (1/10)^∞

Refusing to count the "infinite part" of the sum has funny consequences:

if

0.999...9 = 1, then

0.999...8 = 0.999...9, because we have

0.999...9 = 1 - 1.(1/10)^∞ and

0.999...8 = 1 - 2.(1/10)^∞ but

1.(1/10)^∞ = 0, isn't it? So it must be that

2.(1/10)^∞ = 0. And therefore

0.999...7 = 0.999...8, because ...

And so on. This error demonstration proves that no real number exists, because they all equal a nearby integer. Beautiful.

2013-03-25

So when in base 10, the distance between 2 infinitely close points is: (1/10)^∞. What if we change to base 12 for example? Well, doing the same demonstration, the result should be: [(1/12)^∞] in base 12.

But isn't it that 2 infinitely close real numbers are separated by the same distance, in any base? If true, then we can assume that: [(1/10)^∞] in base 10 = [(1/12)^∞] in base 12, and in the general case:

[(1/a)^∞] in base a = [(1/b)^∞] in base b

But, ∞ = ∞, isn't it? So it must be that: [a] in base a = [b] in base b. [10] in base 10 = [12] in base 12. No it's not. Whoops.

That's why we should never use a result containing "∞" as a part of another demonstration, but always take the result that is true for any "n".

It's not working because we can't write for example:

[0,999...] in base 10 = [0,777...] in base 8

And to prove that is easy: we can't find a "n" where this would be true:

[9.[S4(n, 1/10) - 1]] in base 10 = [7.[S4(n, 1/8) - 1]] in base 8

But this is true for any "n":

[(1/a)^n] in base a - [(1/b)^n] in base b = [(1/LCM(a,b))^n] in base LCM(a,b)

For example:

[1/10^∞] in base 10 = [1/8^∞] in base 8 - [1/40^∞] in base 40

So converting real numbers from one base to another (one scale to another), even with infinitely many digits, loses a part of the numbers if we don't use a well chosen other base, which is LCM of the 2 bases.

2013-03-27

In a way, this error is also related to this assumption (x non null):

x / ∞ = 0

∞ = x / 0

"x / ∞ = 0" is basically saying that "the smallest distance that we can imagine between 2 real numbers, is 0". It's false. The smallest is not null, otherwise no real number would be different from each other, so they would be all the same, and would not exist.

The smallest distance between 2 real numbers is precisely: 1 / ∞, we can call it "epsilon" or whatever, but it's not null, obviously, otherwise "zero" would be his name.

And "∞ = x / 0" is equivalent for this question "if you give me 0 everyday, how many days do I need to reach the total amount of x" to answer "infinitely many". It's also false, and stupid. If you receive 0 everyday, you will never reach x.

Hilbert's paradox of the Grand Hotel

2013-03-25

Finitely many new guests:

"Suppose a new guest arrives and wishes to be accommodated in the hotel. Because the hotel has infinitely many rooms", we can shift to the right every client of every room, and free the number of rooms needed to fit the new guests.

We shift to the right every client of the hotel:

hotel(n) = list(room(x) for x from 1 to n)

hotel(n) = list(room(x) for x from 2 to (n+1))

hotel(n) = hotel(for x from 2 to n) + room(n+1)

hotel(∞) = hotel(for x from 2 to ∞) + room(∞+1)

But the "n+1" room does not exists, and what's not true for "n" won't be true to "∞". So this shift of the infinite list to the right is not valid, and we can't add even 1 room with this trick, even if the hotel has an infinite number of rooms.

A way to demonstrate that assumption "we can't add to an infinite list" discuss the elements of the list: let's say that there is a possibility to add 1 element to that infinite list. But, because the first infinite list already encompass everything, how is it possible that the 1 new elements was not in the list? Isn't it that this 1 element was out of "everything" before being added? If true, then the first list did not contain everything and therefore, was not infinite. But we supposed it was, so it must be that adding into it is not possible.

Let's take a concrete example:

let's build a hotel of all the fractions between 0 and 1, there are infinitely many of them:

hotel(n) = list((x / n) for x from 1 to (n-1))

hotel(∞) = list((x / ∞) for x from 1 to (∞-1))

now if we shift this infinite list 1 to the right, we get:

hotel(n) = list((x / n) for x from 2 to (n))

hotel(n) = list((x / n) for x from 2 to (n-1)) + room(n)

hotel(∞) = list((x / ∞) for x from 2 to (∞-1)) + room(n)

we don't have any room free before "n", and we want room 1 to be free, but:

room(n) = n/n = 1

1 is disallowed in our hotel, because it's not a value between 0 and 1. So the conclusion is:

in this Grand full Hotel with an infinite number of rooms, we can't add any new guest, because no room is free

2013-03-26

Another concrete example:

another infinite hotel have all the integers as rooms numbers. Can we add 1 room? Obviously not, otherwise, which one? Every single number is already taken.

And, in general:

we build a hotel with rooms numbers as:

all squares of all the integers,

or all the fractions,

or all the fractions plus all the irrational numbers as well,

or whatever infinite list as big as we can imagine,

then we rename all the rooms with their rank when we count them: 1, 2, 3 ... to infinity.

We can do it because we have enough integers (infinity) to count them all.

So, in this particular hotel with the renamed rooms, if the hotel is full, no new room can be added. And because in every different infinite hotel we can rename the rooms to the integers, the different hotels are all equivalent to this demonstration, so this result is the only possible result.

Where does the error comes from?

This statement:

Imagine that you have "∞" rooms. "∞ + 1" = "∞", so we can free a room on the left, add 1 room on the right, we still will have infinite places on the right of the list.

is false because if we write with a "n" and take "n" to infinity so that the list never ends, then by definition there is nothing "at the right of infinity" in the list: infinity is not a value, and the list only contains values, so infinity does not exists in the list, therefore it has no "right", and everything in the list is on its "left". We only call a list "infinity" when everything was added into it.

It's also false because if we write to "n", then "n+1" is disallowed, and we can't push "n" to infinity in a non-valid statement.

Hilbert's paradox of the Grand Hotel:

we have a Grand Hotel with infinite rooms,

all the rooms are occupied, the hotel is full,

then a new guest arrives and asks for a room.

Question:

Can he fit the hotel?

No, because the hotel is already full, as stated. It's infinitely full.

Now what if another hotel exists,

with the desired number of free rooms, regarding the number of new guests (finite or infinite):

Can the hotel buy the second one and add the new rooms? Yes it can.

And can this trick fit the new guests (finite or infinite)? Yes it does.

And after this extension of the hotel, how many rooms are there? Still an infinity.

But is it a different hotel, or the same? It's a different hotel.

so we still can't fit any new guest in an infinite full hotel,

but we can change the hotel to add free rooms to fit the guests.

For example, if we have a full hotel of all the fractions between 0 and 1, and then we buy another infinite hotel which is free (ready to fit in new guests), it can be all the fractions between 1 and 2.

Now, what are the numbers of the added rooms?

Because all the rooms already have numbers, and they already cover all integers, so that there is no new number that can be used.

This is the place where the trick used by Hilbert is interesting: we can shift all the rooms numbers.

Notice that we don't shift the rooms or the clients, just the numbers.

In fact, we have changed the hotel, so we can count again the rooms and give them their rank in the counting process, as during the first count. We don't need to find a complicated equation, as described in "Infinitely many new guests", "Infinitely many coaches with infinitely many guests each", "Further layers of infinity", ...

So there's no paradox here, but an error.

So, now that we understand where are the errors, let's build a valid paradox(?):

we have a Grand Hotel with infinite rooms,

with each room having a different integer as number of the room,

then we build 1 new room. How can we give it a new number?

=> shift the numbers 1 to the right and give "1" to the new room

then we build "∞" new rooms,

=> shift the numbers "n" to the right, taking "n" to infinity, or use a complicated formula that frees enough numbers (for example "x2" frees all odd numbers)

then we build "∞²" new rooms,

=> shift the numbers "n²" to the right, taking "n" to infinity, or use a more complicated formula that frees enough numbers

then we build "∞^∞" new rooms,

=> shift the numbers "n^n" to the right, taking "n" to infinity, or use an even more complicated formula that frees enough numbers

and so on.

Cantor's diagonal argument

2013-03-26

"Cantor considered the set T of all infinite sequences of binary digits. He constructs the sequence s by choosing its ith digit as complementary to the ith digit of s(i), for every i. By construction, s differs from each s(n), since their ith digits differ. Hence, s cannot occur in the enumeration."

Or so, let's try with a small finite example (in base 2, with n=3):

s(1) = 000

s(2) = 001

s(3) = 010

s(4) = 011

s(5) = 100

s(6) = 101

s(7) = 110

s(8) = 111

So, what is this new "s" that the diagonal can create, which is not in the list? And where is the diagonal? Well, we don't see it, because the number of rows is 2^3 = 8 is greater than the number of digits: 3.

The diagonal does not exists here in this example, and in general also, in base "b" with "n" digits, the number of rows b^n is always greater than the number of digits "n" (a base can't have less that 2 different digits, so b >= 2), so that the diagonal never exists.

In fact, this is precisely the goal of writing in a base: being able to write everything with fewer digits than the number of elements to display.

So, Cantor's diagonal argument is invalid for any "n" and therefore, it can't be used to infinity.

Infinity is not a place, it's a direction. So, saying that something "is true to infinity" is equivalent to saying that it's "true for any n".

Assuming that the diagonal does exist

because we have an infinite number of digits as well as an infinite number of rows, is the same mistake than considering that

a running Achille and a slower tortoise (starting a race at the same point) will eventually meet towards infinity, or change their order in the race.

It's not only a mistake, it's also stupid.

(the number of rows always "run faster" than the number of digits : b^n > n)

How many integers numbers, fractions, irrational numbers, real numbers, etc?

2013-03-26

If we assume that we can write any real numbers in base "b", even with an infinite number of digits, then the smallest real that can be written (with "n" digits) is:

[epsilon(n digits)] in base b = [0.000... 1] in base b

[epsilon(n digits)] in base b = [(1/b)^n] in base b

epsilon(n digits) = (1/10)^n

So the total number of real numbers between 0 and 1 is:

count(n digits) = 1 / epsilon(n)

count(n digits) = 1 / (1/10)^n

count(n digits) =10^n

count(∞ digits) =10^∞

(...)

1. Infinity does not exist as a value, so stop using it as part of a demonstration, or your demonstration will be invalid (and your result, probably wrong). Mathematics can only manipulate values.

2. Writing the limit to infinity should always be a dead end of a demonstration.

3. Search for an equation that is valid for any "n", and in this equation, you'll have an idea of infinity.

Thomas Godart
written from the 2013-03-24
to the 2013-03-27