Timeline for Find $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$ where $f(n)$ is the number of $1$s in $n$'s binary expansion
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 3, 2014 at 15:29 | audit | First posts | |||
| Apr 3, 2014 at 15:29 | |||||
| Mar 22, 2014 at 17:54 | audit | First posts | |||
| Mar 22, 2014 at 18:12 | |||||
| Mar 16, 2014 at 17:04 | vote | accept | Dmitri Kovalenko | ||
| Mar 16, 2014 at 14:38 | history | edited | WimC | CC BY-SA 3.0 | added 405 characters in body |
| Mar 16, 2014 at 12:00 | comment | added | Dmitri Kovalenko | $\sum_{i=1}^\infty \frac {f(n)} {n \cdot (n+1)}= \sum_{i=1}^\infty \frac {f(2n)} {(2n) \cdot (2n+1)} + \sum_{i=0}^\infty \frac {f(2n+1)} {(2n+1) \cdot (2n+2)} = \sum_{i=1}^\infty \frac {f(n)} {(2n) \cdot (2n+1)} +\sum_{i=0}^\infty \frac {f(n)+1} {(2n+1) \cdot (2n+2)}$ Grouping these terms led me nowhere. Would you give me some extra pointers on how get to your final expression and also why natural logarithm happened to appear in the formula? | |
| Mar 16, 2014 at 9:48 | history | answered | WimC | CC BY-SA 3.0 |