I was talking to my math teacher about the classical problem where the king loses a chess match against a peasent and he is asked to give him $2^n$ grains for the $n$'th square (where the squares are numbered $1$ through $64$).This seems to be a classical example of the power of exponentiation.
However my teacher tells me there is a reversal of this problem where the king tells him that he will go further and pretend as if the board was infinite. The peasent agrees being very greedy, but apparently this is bad for the peasent? I am confused here, did I understand incorrectly, why is this good for the King?
If the field-to-field factor is $x$ (instead of $2$), the total number of grains for $n$ fields is $\frac{x^{n+1}-1}{x-1}$. If $|x|<1$ and we let $n$ grow to infinity, this value approaches $\frac1{1-x}$. So if we abuse(!) this limit result and apply it also for the case at hand, i.e. with $x=2$, then there should be $\frac1{1-2}=-1$ grains on an infinite chess board - the peasent owes the king one grain.
Another reason to justify the result: If we assume all grains placed and then double the content of each field, and then prepend the infinite board with a single field with one grain on it, we obtain the original situation. Hence the number of grains must be a solution of $2\cdot N+1=N$. The only finite solution is indeed $N=-1$.