Counting
how to sum if the terms do not decrease
The sums of infinite sets of numbers (series) can have a finite value. The main thing is that the numbers in such a series decrease; the farther from the beginning of the series - the number should be less. Moreover, the numbers should decrease quickly enough. For instance, the sum
diverges (that is, it does not have a final value).
Fig.1. Sum of series 1/n grow up to infinity with number of terms.
Whereas the sum
converges to the finite value 2 (the more terms we take, the closer the result is to 2).ย
Fig.2. The sum of the series 1/2n vs the number of terms 1 to 10ย
For series where the numbers do not decrease, the situation is ... shall we say: strange. First, let's look at a couple of examples where the numbers in the series change sign (+/-) from term to term.
If we take one, three, five ... - any odd number of terms, then, obviously, the sum will be equal to 1. If we take two, four ... - an even number of terms, then we will get 0 in the sum. What about infinite number of terms, is this number even or odd ๐ค? What will be the sum of an infinite number of these terms? Obviously not infinity. So the finite value? How many?
Let's denote this value (the sum of an infinite number of terms) S. Now let's write down what happens when subtracting 1-S :ย
So after all, this again turns out to be an infinite sum of ones with alternating signs! That is, S. So,
Or
Or
The result is unexpected and strange. One can, of course, say that this is the arithmetic mean of the sums for an even number of terms (0) and for an odd number of terms (1). But that doesn't prove anything...
This sum becomes either negative or positive (for an even or odd number of terms) and, at the same time, the sum grows in absolute value.
Fig.3.The sum vs the number of terms 1 to 50
It is hard to assume that an infinite number of terms will give a finite sum ... However, let's try to guess. (Above - combining the terms in different ways - we got three different results for this sum. Here we will try "a new way".)
Again, let's denote
Now let's take two such sums, add zero in front of one sum (what, of course, will not change the sum); and, finally, let's add these two sums to each other term by term:ย
In example 1, we got that the sum of 1-1+1-1โฆ is equal to ยฝ.ย That is, in our case, adding the two sums, we get:ย
Or
It also looks strange for the sum of terms which are increasing in absolute value.
Howeverโฆ
Nobody promised that it would be easy with infinities ๐.ย
The great Norwegian mathematician Niels Henrik Abel (1802-1829) called divergent series an invention of the devil. In order to somehow work with divergent series without tricks, Abel came up with an interesting general approach.
First, let's recall that almost any (smooth) function can be represented as an infinite polynomial (Taylor/Maclaurin series).
Abel's idea - simplified - is this: if we consider a series of numbers as coefficients of a Taylor series (an) of some function f(x), then the sum of this series of numbers is the value of the function f at x=1.
Quite logical. And by the way, the series from examples 1 and 2 correspond to fairly simple functions:
And, for x=1, these functions are equal to ยฝ and ยผ, respectively (!)
It is interesting to see how the partial sums (sums of the first few terms) of polynomials behave depending on the value of x. And one more thing - to compare them with the function to which the series converges.
Fig.4. Dependences of the partial sums of the first 3, 4, 5, 10, 15 terms of the series and the function 1/(1+x) on the argument xย
Fig.5. Dependences of the partial sums of the first 3, 4, 5, 10, 15 terms of the series and the function 1/(1+x)2 on the argument x
It can be seen that at start (near to X=0) all partial sums practically coincide with the corresponding function (as expected, as it should be). As X increases, the values of the partial sums become less accurate (they move further and further away from the function). Moreover, the more terms in the partial sum, the โwiderโ the area where the partial sum (its value) practically coincides with the function. But outside this region, for large X, the difference between the values of the partial sum and the function grows more and more. And the sign of this difference depends on the (even or odd) number of terms in the partial sum. Each subsequent term (of the form aXn) โtrying to pullโ the sum to the function and adds or subtracts less and less near X = 0 and more and more as X increases (near X = 1). Partial sums "wag their tail" more and more above and below the function.
Such an unstable convergence ...
Other methods/approaches also have been developed to determine the sums of infinite non-decreasing sequences/series. But to discuss them, we need to "dive" deeper into mathematics, get familiar with a specific of zeta- function ... So, not here. There are special textbooks for this.
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Here is another interesting example. Is it possible to find the sum of all natural numbers? Now the sign of the terms does not change, and the terms themselves increase ... to infinity. Let's try to find the sum?๐ค
Then
Let's find the difference by subtracting term by term as shown below:ย
Recall (example 2 above) that 1-2+3-4โฆ = ยผ, then we get:
That is, the sum of all natural numbers is
Wow!? ๐ฎ
This result got by Srinivasa Ramanujan Aiyangar (1887โ1920).
https://en.wikipedia.org/wiki/Ramanujan_summation
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Let's consider
Then add 0 (in front):
Now subtract one from another term by term:
Is the sum of an infinite number of units equal to zero? ๐ค
Let's repeat the trick again - now with this strange result. Add 0 in front again and subtract term by term from the sum of an infinite number of units:ย
That is, 0 = 1. Are you serious? ๐ฎ๐
So, let's repeat once again: infinities must be handled carefully, with great care and attention. The main thing is not to forget what is behind the infinity in each specific task.
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As for tricks with permutations of terms, with term-by-term subtraction / addition, they should be trusted with caution. Not everything familiar is true when it comes to infinityโฆ