Reverse calculation of the sum of a special series (sum of reciprocals)
Author: kimioChanExecutions: 2474
※ This library was created by a member. This site is not responsible for its content.
| n = |
No customer feedback for this calculation yet.
This page has been automatically translated by machine translation. Please note that translations may not be entirely accurate.
※ This library was created by a member. This site is not responsible for its content.
| n = |
No customer feedback for this calculation yet.
Given the sum of reciprocals Sn = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/n = Σ(1/k), (k = 1, ..., n),
find the value of n that does not exceed Sn.