Verified answer

The sum of the series 1+2+3+.....is equal to ∞.

You subtracted ∞ from ∞ and put down it's value as 0.It is actually an indeterminate value.....

Your manipulation of the second line of your equation seemed not in order. If I may say, my esteemed colleague, sir, you're like railroading the result (hehe).

If I may venture my own manipulation (might be obsolete!), this is how it goes:

...1 + 2 + 3 + 4 +...= .......1 + 2 +3 + 4 +...<----- eq 1

...-1 - 2 - 3 - 4 -......= - 1 - 2 - 3 - 4..... <-----eq 2 (why can't I choose which to subtract?)

===========================

0 + 0 + 0 + 0 = -1 - 1 - 1 - 1 ......

0 = - ∞

..... and you have stated that -------> 0 = ∞

0 = 0 (identity)

0 = ∞ (yours)

0 = - ∞ (mine)

therefore:

∞ = - ∞

....and rightly so.....

These quantities should be equal because this is supported in Astronomy by the Principle of Homogenity of Space, namely, that wherever we look, space is the same, and that whichever way it goes, it will terminate in INFINITY and so positive equals negative, and everything becomes zero as the summation of both directions and everything goes back to nothing, just like what it should be before the Big Bang.

never kind of thought that the guy who gave the best answer to my question would ask such an easy question, and by the way santosh was correct inf - inf is an indeterminate form....

....1 + 2 + 3 + 4 +...= 1 + 2 + 3 + 4 +...

...-1 - 2 - 3 - 4 -.......=.....- 1 - 2 - 3 -

this step was like what we used to do to find an arithmetico-geometric progression sum!

by the way rhs is also 0, if the lhs is 0<lol> actually it was kind of funny to subtract inf from inf in lhs and write it a 0.(as 1+2+3+......inf = inf)

1 + 2 + 3 + ..... + n

....-1 - 2 -...........-(n-1) -n

1 + 1 + 1 + ....... +1 - n = 0

sorry for all this trash, you asked santosh not to make it easy so i decided to take a rough way out

I'm pretty sure the statement '1+2+3+... = 1+2+3+... ' is itself fallacious, algebraic manipulation only applies to finite quantities :|

I believe this 'proof' is a demonstration, though somewhat skewed, of the Riemann Rearrangement Theorem at work. Namely, given an infinite number of positive and negative terms, we can create a series through a specific ordering of these terms that converges on any value of our choosing: Positive, negative, finite, or infinite.

-- Also, what happened to your question on convex shapes and bounding squares? I was intrigued.

I got another one~

1 × 2 × 3 × 4 × ... = 1 × 2 × 3 × 4 × ...

Divide both sides from itself, as follows

1/1 × 2/2 × 3/3 × 4/4 × ... = 1/2 × 2/3 × 3/4 × 4/5 × ...

Simplifying it gives, and by cutting out denominators and numerators in pairs

1 × 1 × 1 × 1 × 1 × ... = lim(n→∞) 1/n

This ends up with

1 = 0

What's the fallacious step?

EDIT: haha yeah totally man. My very first fallacies (I've discovered/created myself) is

-1 = (-1)^1 = (-1)^(2/2) = ( (-1)^2 )^(1/2) = 1^(1/2) = 1

I felt a little bit under despair afterwards 'cause I can't tell what's real and what's fake anymore.

look the case which involves 'n' terms

1+ 2+ 3+ 4.............+ n = 1+ 2 + 3 + 4 + ...........+n

-1- 2- 3 -4 ...............-n = -1 - 2 - 3 - -(n-1) - n

0+0+0+0.................+0= 1+ 1+ 1+1.................+1-n

the fallacious step is to forget the last term -n

1+1+1+1+1+1.............-n as n tends to infinity is not infinity it is zero.