3 ms·

But a number between 0 and 1 could be arbitrarily close to 0 and 1, which I thought could be proved to be equal to 0 or 1, which are integers. So isn't this not

• panic 10y ago
  If 0 < a < 1, then a can't be equal to 0 or 1. There's no "arbitrarily close" for numbers, just sequences of numbers (like the partial sums in the series).
  • irishsultan 10y ago
    But in this case the right side is exactly that, the sum of an infinite sequence of numbers (which equals the limit of the partial sum for that sequence if it converges (which this sequence of partial sums does)). The reason it still works is because it's actually 0 < s_0 < a < s_1 < 1, so even if a approaches s_1 arbitrarily close it still won't be equal to an integer (same is true if it goes arbitrarily close to s_0 of course).
• siegelzero 10y ago
  It's a contradiction because there are no integers strictly between 0 and 1.
• nilkn 10y ago
  If the proof were just that the sum converges to some number 0 <= x <= 1, then you'd be right. But it's showing specifically that the sum lies between the first term s_0 (which is strictly less than 1) and the sum of the first two terms s_0 - s_1 (which is strictly greater than 0). So there's no possibility for it to converge to a value outside of the interval [s_0-s_1,s_0], which is itself strictly contained inside [0,1].