Sorry if its already been answered.
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | jiangly | 3670 |
| 3 | Kevin114514 | 3657 |
| 4 | maroonrk | 3355 |
| 5 | Um_nik | 3353 |
| 6 | squareOf105 | 3346 |
| 7 | tourist | 3343 |
| 8 | strapple | 3324 |
| 9 | heuristica | 3322 |
| 10 | jiangbowen | 3291 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 158 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | AmShZ | 143 |
| 5 | Um_nik | 141 |
| 6 | Errichto | 139 |
| 7 | adamant | 137 |
| 8 | BledDest | 134 |
| 9 | maroonrk | 133 |
| 10 | qwexd | 130 |
Sorry if its already been answered.
| Name |
|---|



We can approximate the sum of this infinite harmonic sequence with the natural log function (Think of it like integration instead of addition), and the error is less than $$$1$$$ (It's called the Euler-Mascheroni constant)
lets say it is smaller than
1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/8 + ...(we have1/2^i2^itimes) and(1/2^i) * 2^i = 1so we have n * log(n)Good job bruh
Bibek..rocks
We can apply integral test for it 1+/2+1/3+...+1/n < integral of (1/x)dx with lower limit 1 and upper limit n. So the sum is < log(n)
Natural log function