# Partial Sums of Harmonic Series

Given an positive integer $n$, return the least positive integer $k$ such that the $k$th partial sum of the harmonic series is greater than or equal to $n$. For example, if $n = 2$, then $k = 4$, because $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12} > 2$.

# More Input/Output Examples

```
Input -> Output
1 -> 1
3 -> 11
4 -> 31
```

This is code-golf, so shortest code wins.

Japt, 11 10 bytes >0©Òß …

1y ago

[Python 3], 35 bytes …

2y ago

AppleScript, 178 bytes ``` …

2y ago

[Raku], 27 bytes {+ …

2y ago

[Scala], 55 bytes ( …

2y ago

Stax, 9 bytes Ç≈f♠É↔X+ö …

2y ago

[JavaScript (Node.js)], 47 byt …

2y ago

[Jelly], 8 bytes İ€S<¬ʋ …

1y ago

[J], 23 bytes >:@]^:(>1 …

1y ago

JavaScript, 29 bytes …

1y ago

Goruby, 38 bytes ``` ->n{dw{ …

5mo ago

## 11 answers

# AppleScript, 178 bytes

```
set n to text returned of (display dialog "" default answer "") as number
set k to 0
set h to 0
repeat
set k to k + 1
set h to h + (1 / k)
if h >= n then exit repeat
end repeat
k
```

It's really a shame that AppleScript doesn't support explicit stdin.

# Raku, 27 bytes

```
{+([\+](1 X/1..*)...*>=$_)}
```

```
{ } # Anonymous code block
[\+]( ) # Get the partial sum of
1 X/ # 1 over each of
1..* # All positive integers
...*>=$_ # Take from the list until it is bigger than the input
+( ) # And return the length of the list
```

This will start to suffer precision issues, but you can add `.FatRat`

to work correctly (but much slower)

# Python 3, 35 bytes

```
f=lambda x,i=1:x>0and-~f(x-1/i,i+1)
```

Subtracts each `1/i`

in turn from the initial value until the result is no longer positive.

#### 0 comment threads

# Scala, 55 bytes

```
(Stream from 1 map 1.0./scanLeft.0)(_+_)indexWhere _.<=
```

Written in a more sane manner:

```
n => Stream.from(1).map(1.0/_).scanLeft(0.0)(_+_).indexWhere(_ >= n)
```

Explanation:

```
(
Stream from 1 //Make an infinite list of integers, starting at 1
map 1.0./ //Find the reciprocal of each (divide 1.0)
scanLeft .0)(_+_) //Sum each partial sequence
indexWhere //Find the index where
_.<= //n (underscore) is less than or equal to a partial sum
```

#### 0 comment threads

# JavaScript (Node.js), 47 bytes

```
f=(a,b=0,c=a=>a&&1/a+c(a-1))=>c(b)<a?f(a,b+1):b
```

For large numbers, a rounding error will occur and yield incorrect result. The solution below works for any integer (no rounding errors):

# 93 bytes

```
f=(a,i=0n,p=a=>g(a)/c(a),g=(a,b=a)=>a&&c(b)/a+g(~-a,b),c=a=>a?a*c(~-a):1n)=>p(i)<a?f(a,-~i):i
```

#### 0 comment threads

# Jelly, 8 bytes

```
İ€S<¬ʋ1#
```

## How it works

```
İ€S<¬ʋ1# - Main link. Takes n on the left
ʋ - Group the previous 4 links into a dyad f(k, n):
€ - Over each integer 1 to k:
İ - Get its reciprocal
S - Sum
< - Less than n?
¬ - Logical not
1# - Find the first integer k ≥ n such that f(k, n) is true
```

#### 0 comment threads

# J, 23 bytes

```
>:@]^:(>1#.1%1+i.)^:_&1
```

```
>:@]^:(...)^:_&1 Increments from 1 while
>1#.1%1+i. the harmonic sum up to that is less than the input
```

#### 0 comment threads

# JavaScript, 29 bytes

```
f=(n,x=0)=>n>0?f(n-1/++x,x):x
```

## 0 comment threads