The sum of all natural numbers: 1 + 2 + 3 + 4 + .... Part 2

Many people know that

$$

$$1 + 2 + 3 + \ dots = - \ dfrac {1} {12}.$$

But in reality

$$

$$1 + 2 + 3 + \ dots = - \ dfrac {1} {8}.$$

Let us consider in more detail the first result. Of course, a series of natural numbers diverges in the classical sense (in the sense of convergence of a sequence of partial sums: it, of course, has no limit). In this article, the author mentions other summation methods, such as the Cesaro method and the Abel method. Here are some examples: the sum of such a series

$$

$$\ sum \ limits_ {n \ geqslant 0} (-1) ^ n = 1 - 1 + 1 - 1 + 1 - 1 + \ dots$$

using the cesaro method will be equal $$$\ dfrac12$.

Another example:

$$

$$1 - 2 + 3 - 4 + 5 + \ dots = \ dfrac14.$$

In my opinion, it is wrong to say that the sum of the first row is equal to $$$\ dfrac12$; correctly say that the sum of the first row in the sense of Cesaro is equal to $$$\ dfrac12$. Similarly for the second: its sum in the sense of Abel is equal to $$$\ dfrac14$.

In view of this, in the first result (that $$$- \ dfrac {1} {12}$) there is a substitution of concepts, which leads to a contradiction with common sense.

We now consider in more detail the second result. First, we denote the entire amount for $$$X$:

$$

$$1 + 2 + 3 + 4 + \ dots = X.$$

Now we perform the following transformations:

$$

$$1 + 2 + 3 + 4 + \ dots = 1 + \ underbrace {2 + 3 + 4} _ {9} + \ underbrace {5 + 6 + 7} _ {18} + \ underbrace {8 + 9 + 10} _ {27} + \ dots =$$

$$

$$1 + 9 + 18 + 27 + \ dots = 1 + 9 \ underbrace {\ left (1 + 2 + 3 + \ dots \ right)} _ X = X.$$

From here

$$

$$1 + 9X = X \ Rightarrow X = - \ dfrac {1} {8}.$$

There is another solution. Combine the terms in another way:

$$

$$1 + 2 + \ underbrace {3 + 4 + 5 + 6 + 7} _ {25} + \ underbrace {8 + 9 + 10 + 11 + 12} _ {50} + \ dots =$$

$$

$$= 1 + 2 + 25 \ underbrace {\ left (1 + 2 + 3 + \ dots \ right)} _ X = X,$$

i.e

$$

$$1 + 2 + 25X = X \ Rightarrow X = - \ dfrac {3} {24} = - \ dfrac {1} {8}.$$

In fact, starting from the top three, we can distinguish 7 terms, the sum of which will be 49, and we will come to the equation

$$

$$1 + 2 + 3 + 49X = X,$$

which will give the same result.

In general, you need to act like this: select the first $$$n$terms, and then in parentheses take $$$2n + 1$terms:

$$

$$1 + \ dots + n + \ underbrace {\ left (n + 1 + \ dots + 3n + 1 \ right)} _ {(2n + 1) ^ 2} + \ underbrace {\ left (3n + 2 + \ dots + 5n + 2 \ right)} _ {2 (2n + 1) ^ 2} + \ dots =$$

$$

$$1 + \ dots + n + (2n + 1) ^ 2 \ left (1 + 2 + 3 + \ dots \ right) = X.$$

Arithmetic progression $$$1 + \ dots + n$is equal to $$$\ dfrac {n (n + 1)} {2}$, therefore, we obtain the equation

$$

$$\ dfrac {n (n + 1)} {2} + (2n + 1) ^ 2X = X,$$

where does it turn out that

$$

$$X = - \ dfrac {1} {8}.$$