1/infinity is in fact, not zero.
For the same reason that 1/infinity = 2/infinity so thus, 2 must be zero, right? Mathmatically, this does not hold true.
1/0 is NOT infinity, since it's 0 x infinity will always be zero.
Thus infinity times zero is still, zero.
The one thing you have to remember about summations is that it's an entity of it's own, that rounds to the nearest figure. That's why it's it's own thing.
So 1 + 1/2 + 1/4 + etc. will be 2 up to infinity. But it's actually will not ever truly reach 2. This is called a limit. The limit of the summation of 1 + 1/2 etc. to infinity is 2, but that's becuase it never actually reaches 2. That's why the solution in calculus is called the limit, it's the number it can't reach. So 1/infinity has a limit of 0, but it is not actually zero itself.
Consider though that you already divide by zero. Attempt to do this algorithm; 50/10. Simple, it is 5.
But, the algorithm shows something very clearly.
50
10
You take 0, divided it by zero. Pluasible. But then you divide 5 by zero! What do we do? Well, the 0 is divided by 1, and the 5 is divided by zero. The answer should be undefined; but it's not, it's 5.
This is becuase when you divide that 5 by 0, you get zero. And it doesn't serve as a place holder, etheir; it's literally, nothing. Thus the end equation is not 5, next to a 0, but 5 plus that zero. Thus you get that, 5/0 = 0, and in other situations, we find that when you divide by zero, you get zero. The 5 is eventually divided by that 1, and thus you get 5, since 5/1 is 5. But the place holder of the zero is still, in fact, zero.
Edited by Manoka, 20 May 2013 - 03:47 PM.