# Revision history [back]

One possibility to get around this infinite sumation is to use truncation

sage: sum(1 / (2^x - 1), x, 1, 10).n()
1.60571827189075
sage: sum(1 / (2^x - 1), x, 1, 30).n()
1.60669515148397
sage: sum(1 / (2^x - 1), x, 1, 100).n()
1.60669515241529


Your sum is converging pretty fast. The tail after $n$ is of the order of $2^{-n}$.

One possibility to get around this infinite sumation is to use truncation

sage: sum(1 / (2^x - 1), x, 1, 10).n()
1.60571827189075
sage: sum(1 / (2^x - 1), x, 1, 30).n()
1.60669515148397
sage: sum(1 / (2^x - 1), x, 1, 100).n()
1.60669515241529


Your sum is converging pretty fast. The tail after $n$ is of the order of $2^{-n}$.$2^{-n}$. In particular, since real numbers have a default of 53 bits of precision, the evaluation of the sum is the same at 53 and 100 (but not at 52)

sage: sum(1 / (2^x - 1), x, 1, 52).n() == sum(1 / (2^x - 1), x, 1, 100).n()
False
sage: sum(1 / (2^x - 1), x, 1, 53).n() == sum(1 / (2^x - 1), x, 1, 100).n()
True


The fact that it is exactly 53 in this case is because the size of the tail is exactly $2^{-n}$... for other sums it might be different. But whatever convergent series you are considering, its numerical evaluation will be constant from a certain point.