Non-commutative sum?

Question

I'm attemting to calculate $$\sum_{n=1}^{10000} \frac{1}{n^2}$$ numerically, using a computer program. However, if I sum the terms in the natural order (starting from $n=1$ and going up one unit every time) I get a different result that if I compute the sum in the reversed order (starting from $n=10000$ and going down one unit at a time). I understand it must have something to do with the rounding the computer does in every single calculation but I can't seem to explain the phenomenon in a clear manner. Any help? Thank you in advance.

Answer 1

Perhaps a simpler example may help. Let's say you have a computer that does decimal floating-point arithmetic with only $3$ digits, and you want to add $999 + 1 + 1 + \ldots 1$ (with $100$ 1's).

If you do it in that order, the first addition gives you $999 + 1 = 1000 = 1.00 \times 10^3$, and each subsequent addition leaves it there ($1000 + 1 = 1001$, but rounded to $3$ digits that's still $1.00 \times 10^3$).

On the other hand, try it in the reverse order. $1 + 1 = 2$, $2 + 1 = 3$, \ldots, $99 + 1 = 100$, and finally $100 + 999 = 1099 = 1.10 \times 10^3$.

Answer 2

The rounding (or better said "approximate representation") takes place both in the calculation of the fraction, as well as in the addition of the terms. The latter makes the result dependent on the order of additions.