# Definition

In Mathematics, Harmonic Sequence refers to a sequence where i.e. the nth term of the sequence equals the reciprocal of n.

## Introduction

In this challenge, given a positive integer n as input, output the Partial Sum of first n terms of the Harmonic Sequence.

# Input

You'll be given a positive integer (within the range of numbers supported by your language). It can be either of Signed and Unsigned (depends on you), since the challenge requires only positive integers.

You can take the input in any way except assuming it to be present in a predefined variable. Reading from file, terminal, modal window (prompt() in JavaScript) etc. is allowed. Taking the input as function argument is allowed as well.

# Output

Your program should output the sum of the first n terms of the Harmonic Sequence as a float (or integer if the output is evenly divisible by 1) with precision of 5 significant figures, where n refers to the input. To convey the same in Mathematical jargon, you need to compute where n refers to the input.

You can output in any way except writing the output to a variable. Writing to screen, terminal, file, modal window (alert() in JavaScript) etc. is allowed. Outputting as function return value is allowed as well.

# Test Cases

The Test Cases assume the input to be 1-indexed

Input     Output
1         1
2         1.5
3         1.8333
4         2.0833
5         2.2833


# Winning Criterion

This is , so the shortest code in bytes wins!

• Could you give us some testcases? – user41805 May 28 '17 at 9:50
• What precision is required? Exact output is generally only possible as a fraction, but in many languages that will have to be separate numbers for numerator and denominator. Can we output a)a float, b)a fraction or integer pair c)either? – Level River St May 28 '17 at 9:52
• @Arjun The harmonic series grows to infinity so it will get hard to meet 10 decimal places as the number gets into the thousands and millions. I would go for significant figures rather than decimal places, and I see no need to be so precise. 5 significant figures should be enough. so 9.9999E10 rather than 99999999999.9999999999 – Level River St May 28 '17 at 10:17
• Can we go over 5 significant figures? – Erik the Outgolfer May 28 '17 at 10:25
• By the way, it's known that the harmonic sequence does not contain any integers other than the initial a_1 = 1. (Idea of proof that a_n is not an integer for n>1: let 2^k be the largest power of 2 not exceeding n; then 2^k divides the denominator of a_n.) – Greg Martin May 28 '17 at 19:53

# Jelly, 3 bytes

İ€S


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1-indexed.

Explanation:

İ€S Main link, monadic
İ€         1 / each one of [1..n]
S Sum of


# Python 3, 27 bytes

h=lambda n:n and 1/n+h(n-1)

• 0-indexing or 1-indexing? – Arjun May 28 '17 at 11:29
• Throws RuntimeError when handling input greater than the recursion limit, 1000 by default. – sagiksp May 28 '17 at 15:04
• you can do sys.setrecursionlimit(473755252663) but the stack will eventually overflow quite easily – cat May 29 '17 at 4:31
• @Arjun it's 1-indexed – shooqie May 29 '17 at 20:23

# JavaScript, 19 18 bytes

1 byte saved thanks to @RickHitchcock

f=a=>a&&1/a+f(--a)


This is 1-indexed.

f=a=>a&&1/a+f(--a)

for(i=0;++i<10;)console.log(f(i))

• From what I've seen of other posts, you can remove f= from your answer to save 2 bytes. – Rick Hitchcock May 28 '17 at 10:44
• @RickHitchcock I cannot remove f= because the function is recursive and it references itself in f(--a). But if this was not a recursive solution, I would have been able to do that – user41805 May 28 '17 at 10:45
• Ah, makes sense! Save one byte with f=a=>a&&1/a+f(--a). – Rick Hitchcock May 28 '17 at 10:49
• @RickHitchcock Nice one! – user41805 May 28 '17 at 10:52

# APL (Dyalog), 5 bytes

+/÷∘⍳


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You can add ⎕PP←{number} to the header to change the precision to {number}.

This is 1-indexed.

### Explanation

+/÷∘⍳                     Right argument; n
⍳                     Range; 1 2 ... n
÷                       Reciprocal; 1/1 1/2 ... 1/n
+/                        Sum; 1/1 + 1/2 + ... + 1/n


# Mathematica, 2120 16 bytes

This solution is 1-indexed.

Sum[1./i,{i,#}]&

• It is 1-indexing – ZaMoC May 28 '17 at 11:31
• > You must not use a built-in to calculate the partial sum of the first n elements. (Yeah, it's for you Mathematica!) – MCCCS May 28 '17 at 11:32
• OP means I can't use HarmonicNumber[#] & – ZaMoC May 28 '17 at 11:35
• And one can further shorten to Tr[1./Range@#]&. – Greg Martin May 28 '17 at 19:56
• @Ian Mathematica may display 5 sig fig, but the function returns machine-precision numbers (52 binary bits or just under 16 decimal digits of precision) – LLlAMnYP May 30 '17 at 9:12

# PHP, 33 Bytes

1-indexing

for(;$i++<$argn;)$s+=1/$i;echo$s;  Try it online! # Pari/GP, 18 bytes n->sum(i=1,n,1./i)  1-indexing. Try it online! # CJam, 11 bytes 1.ri,:)f/:+  Try it online! 1-indexed. # Japt-x, 865 3 bytes õpJ  With some thanks to ETHproductions Try it online • 0-indexing or 1-indexing? – Arjun May 28 '17 at 11:27 • I think you can save a byte with õ x@1/X – ETHproductions May 28 '17 at 12:25 • ...and another couple bytes by using XpJ instead of 1/X :-) – ETHproductions May 28 '17 at 12:27 • Thanks, @ETHproductions :) I twigged those as soon as I walked away. – Shaggy May 28 '17 at 12:45 • Actually I don't think you even need the _ due to auto-functions. I should really write that tip :P (I should have time today or tomorrow, due to it being Memorial Day) – ETHproductions May 28 '17 at 14:50 # CJam, 11 10 bytes 1 byte removed thanks to Erik the outgolfer ri),{W#+}*  This uses 1-based indexing. Try it online! ### Explanation ri e# Read integer, n ) e# Increment by 1: gives n+1 , e# Range: gives [0 1 2 ... n] { }* e# Fold this block over the array W# e# Inverse of a number + e# Add two numbers  • You can use W instead of -1. – Erik the Outgolfer May 28 '17 at 12:24 • @EriktheOutgolfer has outgolfed himself :-) – Luis Mendo May 28 '17 at 13:04 • @LuisMendo I like my name, it's just a name. And yes I outgolfed myself in the process of helping a fellow golfer golf even further. – Erik the Outgolfer May 28 '17 at 13:10 • @Erik It was meant as a joke. Thanks for the help – Luis Mendo May 28 '17 at 13:57 ## Haskell, 20 bytes f 0=0 f n=1/n+f(n-1)  ## Original solution, 22 bytes f n=sum[1/k|k<-[1..n]]  These solutios assumes 1-indexed input. # R, 15 bytes sum(1/1:scan())  Try it online! ## Tcl 38 bytes proc h x {expr$x?1./($x)+\[h$x-1]:0}


That's a very dirty hack, and the recursive calls pass literal strings like "5-1-1-1..." until it evaluates to 0.

• Thanks @Christopher for the formatting. With that, the duplication of backslash was no longer necessary. – avl42 May 28 '17 at 20:30
• No problem! It looks better – Christopher May 28 '17 at 21:30

# 05AB1E, 3 bytes

LzO


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1-indexed.

# MATL, 5 bytes

:l_^s


This solution uses 1-based indexing.

Try it at MATL Online

Explanation

    % Implicitly grab input (N)
:   % Create an array from [1...N]
l_^ % Raise all elements to the -1 power (take the inverse of each)
s   % Sum all values in the array and implicitly display the result


# Axiom, 45 34 bytes

f(x:PI):Any==sum(1./n,n=1..x)::Any


1-Indexed; It has argument one positive integer(PI) and return "Any" that the sys convert (or not convert) to the type useful for next function arg (at last it seems so seeing below examples)

(25) -> [[i,f(i)] for i in 1..9]
(25)
[[1,1.0], [2,1.5], [3,1.8333333333 333333333], [4,2.0833333333 333333333],
[5,2.2833333333 333333333], [6,2.45], [7,2.5928571428 571428572],
[8,2.7178571428 571428572], [9,2.8289682539 682539683]]
Type: List List Any
(26) -> f(3000)
(26)  8.5837498899 591871142
Type: Union(Expression Float,...)
(27) -> f(300000)
(27)  13.1887550852 056117
Type: Union(Expression Float,...)
(29) -> f(45)^2
(29)  19.3155689383 88117644
Type: Expression Float


# Pyth, 5 bytes

scL1S


Try it here.

1-indexed.

# Husk, 3 bytes

ṁ\ḣ


Try it online!

Output is a fraction.

# C, 54 bytes

i;float f(n){float s;for(i=n+1;--i;s+=1./i);return s;}


Uses 1-indexed numbers.

## Haskell, 21 bytes

f n=sum$map(1/)[1..n]  # Brachylog, 6 bytes ⟦₁/₁ᵐ+  Try it online! This is 1-indexed. ### Explanation ⟦₁ Range [1, …, Input] ᵐ Map: /₁ Inverse + Sum  ## QBIC, 13 bytes [:|c=c+1/a]?c  Explanation [ | FOR a = 1 to : the input n c=c+ Add to c (starts off as 0) 1/a the reciprocal of the loop iterator ] NEXT ?c PRINT c  # Gol><>, 8 bytes F1LP,+|B  Try it online! ### Example full program & How it works 1AGIE;GN F1LP,+|B 1AGIE;GN 1AG Register row 1 as function G IE; Take input as int, halt if EOF GN Call G and print the result as number Repeat indefinitely F1LP,+|B F | Repeat n times... 1LP, Compute 1 / (loop counter + 1) + Add B Return  # Rust, 36 bytes |n|(1..=n).map(|n|1./n as f64).sum()  Try it online! Input is an integer, if the input is <=0 the output is zero, and inputting one gets one. # C (gcc), 35 bytes float f(n){return n?1./n+f(--n):0;}  Try it online! # Braingolf, 20 bytes [non-competing] VR1-1[1,!/M,$_1+]v&+


This won't actually work due to braingolf's inability to work with floats, however the logic is correct.

### Explanation:

VR1-1[1,!/M,$_1+]v&+ Implicit input VR Create new stack and return to main stack 1- Decrement input 1 Push 1 [..........] Loop, always runs once, then decrements first item on stack at ] Breaks out of loop if first item on stack reaches 0 1,!/ Push 1, swap last 2 values, and divide without popping Original values are kept on stack, and result of division is pushed M,$_         Move result of division to next stack, then swap last 2 items and
Silently pop last item (1)
1+       Increment last item on stack
v&+   Move to next stack, sum entire stack
Implicit output of last item on current stack


Here's a modified interpreter that supports floats. First argument is input.

# Tcl, 61 bytes

proc h {x s\ 0} {time {set s [expr $s+1./[incr i]]}$x;set s}


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# GolfScript, 17 bytes

~),{2.-1??./\/+}*


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1-indexed

~                   # Parse n to an integer
),                 # Make an array from 0 to n
{           }*   # For each number in the array, skipping the 0
2.-1??          # 2^(2^-1) = sqrt( 2 )  We just need any float
./        # Divide previous number by itself, this will result in the float 1.0
\/      # Divide 1.0 by the current number of the array
+     # Add the result


Without the convertion to float it could be reduced to 10 bytes, but the output would be a fraction.

~),{-1?+}*


Try it online!