hey there ,I have found something:

let S=1+2+3+4....

so S-1=(2+3+4+5)+(6+7+8+9)+(10+11+12+13).

S-1=2(7) +2(15)+ 2(23)...

S-1=2(7)+2(2(7)+1) +2(3(7)+2)...

(S-1)/2 = 2(7) +3(7) +4(7)... +1+2+3+4...

(S-1)/2=7(1+2+3+4...)+1+2+3+4...

(S-1)/2=7S+S

(S-1)=16S

so S=-1/15

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Infinite sums are not commutative. Commutative means for all real nos. a,b we have a+b=b+a. If you keep iterating you will get commutativity for finite sums. Thus if you have a_1+a_2+a_3+...+a_n, for finite n, then you you can change the order of summation. This is not true for the infinite case.

When you replace 1+2+3+4+... with S you are assuming S is a finite sum - which it clearly is not.

The equation: (infinity - 1)/2 = 16*infinity is indeterminate as it leads to infinity - infinity = 0.There are cases where an indeterminate form can have a finite value - but not here. Instead of groups of 4 you could have grouped then in groups of 5,6,7,.. and you would have yielded a different answer each time.

You are using a method like:

S = 1/2 + 1/4 + 1/8 + +1/16 + ....

2S = 1 + 1/2 + 1/4 + + 1/8 + 1/16 + ...

Subtracting term by term gives:

2S - S = 1

S = 1

This technique only works because S itself is a finite number so 2S - S is also finite.

If an infinite series is to have a finite sum then, by definition, we must look at the sequence of "partial sums" S1, S2, S3, S4, ... If this sequence has a limit then we assign that limit to the infinite series.

S1 = 1/2

S2 = 3/4

S3 = 7/8

S4 = 15/16

so S(n) = (2^n - 1)/2^n = 1 - (1/2^n)

as n---> infinity, 1/2^n ---->0

So S(infinity) = 1

In your case the sequence of partial sums clearly blows up. So, in the end, assigning an infinite sum to a variable is not going to get you anywhere.

This is same as sum of Natural numbers by Ramanujan's was -1/12, which is absolutely wrong. Problem is with the infinite series, you cannot say something very clearly in general things.