Just take to account that that S = 2(a+b+c+d+...) does not have a sum.
You wanted to say that the addition of infinitely many positive values doesn't have a sum, right? Well, it kind of make sense, coz the process of addition never stops and therefore the sum changes with each added number. In that case, how can anyone compare the ever-changing result with another number, such as a constant?
You are living under the impression that there is no sum, and that’s because you would define the task of adding the series of positive integers this way
S = 1 + 2 + 3 + 4 + ...
as it can be seen from the way you did it in S = 2(a + b + c + d + ...)
That addition really doesn't have any sum, coz one important element is missing. The instruction to sum the series of positive integers must look like this
S = 1 + 2 + 3 + 4 + ... + n-1 + n
where n → ∞.
The number ‘n’ is important, coz it is defined: n = "a value that approaches infinity."
You may disagree and point out that the definition of 'n' is useless, coz it would take an infinite amount of time to complete the addition -- the process would never end, and therefore "there is no sum."
Actually even the task of adding the first hundred numbers could make someone really impatient . . .
Another famous story has it that in primary school after the young Gauss misbehaved, his teacher, J.G. Büttner, gave him a task add a list of integers in arithmetic progression; as the story is most often told, these were the numbers from 1 to 100. The young Gauss reputedly produced the correct answer within seconds, to the astonishment of his teacher and his assistant Martin Bartels.
Gauss's presumed method was to realize that pairwise addition of terms from opposite ends of the list yielded identical intermediate sums1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on, for a total sum of 50 × 101 = 5050.
And so, due to little Carl and others before him, the infinite addition of the series of whole numbers does have a sum:
S = 1 + 2 + 3 + 4 + ... + n-1 + n = (n
2 + n)/2.
So does any infinite summation.
When an infinite series converges, the sum includes ‘n’ as well and so the algebraic term can be compared with other terms that include ‘n’. But an arbitrary constant doesn’t include it, and so there is “strictly speaking” no way that the result of the summation would ever equal a constant. But remember that there are many ways to skin the cat in math. Here is an infinite convergent series whose sum is Pi.
∞
4
∑ (-1
k)/(2k +1) = Pi
k=0
π (sometimes written pi) is a mathematical constant whose value is the ratio of any circle's circumference to its diameter in Euclidean space; this is the same value as the ratio of a circle's area to the square of its radius. It is approximately equal to 3.14159265 in the usual decimal notation.
Pi can be also expressed in an exact form, such as Pi = c/d (circle circumference/diameter). It all depends on the skill of the mathematician to convert a sum of a convergent series into an exact form and do the comparison with a constant. Sometimes it is possible, more often it is not – it all depends on the particular problem to be solved.
So saying that an infinite series doesn’t have a sum is not true, as any high school kid would confirm so. But they don’t teach OM in the schools, and in this field anything seems to be possible and at the same time impossible.
S = 1 + 2 + 3 + 9 + 24 + 76 + 236 + ... + n-1 + n. (n → ∞)
How does the summation formula look like, Doron?
