Definition

In Mathematics, Harmonic Sequence refers to a sequence where

$$a_n = \frac 1 n$$

i.e. the $$\n_{th}\$$ term of the sequence equals the reciprocal of $$\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

$$\sum_{i=1}^n \frac 1 i$$

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? Commented May 28, 2017 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? Commented May 28, 2017 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 Commented May 28, 2017 at 10:17
• Can we go over 5 significant figures? Commented May 28, 2017 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.) Commented May 28, 2017 at 19:53

Python 3, 27 bytes

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

• 0-indexing or 1-indexing? Commented May 28, 2017 at 11:29
• Throws RuntimeError when handling input greater than the recursion limit, 1000 by default. Commented May 28, 2017 at 15:04
• you can do sys.setrecursionlimit(473755252663) but the stack will eventually overflow quite easily
– cat
Commented May 29, 2017 at 4:31
• @Arjun it's 1-indexed Commented May 29, 2017 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. Commented May 28, 2017 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 Commented May 28, 2017 at 10:45
• Ah, makes sense! Save one byte with f=a=>a&&1/a+f(--a). Commented May 28, 2017 at 10:49
• @RickHitchcock Nice one! Commented May 28, 2017 at 10:52

APL (Dyalog), 5 bytes

+/÷∘⍳


Try it online!

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 Commented May 28, 2017 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!) Commented May 28, 2017 at 11:32
• OP means I can't use HarmonicNumber[#] & Commented May 28, 2017 at 11:35
• And one can further shorten to Tr[1./Range@#]&. Commented May 28, 2017 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) Commented May 30, 2017 at 9:12

PHP, 33 Bytes

1-indexing

for(;$i++<$argn;)$s+=1/$i;echo$s;  Try it online! Pari/GP, 18 bytes n->sum(i=1,n,1./i)  1-indexing. Try it online! CJam, 11 bytes 1.ri,:)f/:+  Try it online! 1-indexed. Japt-x, 865 3 bytes õpJ  With some thanks to ETHproductions Try it online • 0-indexing or 1-indexing? Commented May 28, 2017 at 11:27 • I think you can save a byte with õ x@1/X Commented May 28, 2017 at 12:25 • ...and another couple bytes by using XpJ instead of 1/X :-) Commented May 28, 2017 at 12:27 • Thanks, @ETHproductions :) I twigged those as soon as I walked away. Commented May 28, 2017 at 12:45 • Actually I don't think you even need the _ due to auto-functions. I should really write that tip :P (I should have time today or tomorrow, due to it being Memorial Day) Commented May 28, 2017 at 14:50 Jelly, 3 bytes İ€S  Try it online! 1-indexed. Explanation: İ€S Main link, monadic İ€ 1 / each one of [1..n] S Sum of  CJam, 11 10 bytes 1 byte removed thanks to Erik the outgolfer ri),{W#+}*  This uses 1-based indexing. Try it online! Explanation ri e# Read integer, n ) e# Increment by 1: gives n+1 , e# Range: gives [0 1 2 ... n] { }* e# Fold this block over the array W# e# Inverse of a number + e# Add two numbers  • You can use W instead of -1. Commented May 28, 2017 at 12:24 • @EriktheOutgolfer has outgolfed himself :-) Commented May 28, 2017 at 13:04 • @LuisMendo I like my name, it's just a name. And yes I outgolfed myself in the process of helping a fellow golfer golf even further. Commented May 28, 2017 at 13:10 • @Erik It was meant as a joke. Thanks for the help Commented May 28, 2017 at 13:57 Haskell, 20 bytes f 0=0 f n=1/n+f(n-1)  Original solution, 22 bytes f n=sum[1/k|k<-[1..n]]  These solutios assumes 1-indexed input. R, 15 bytes sum(1/1:scan())  Try it online! Tcl 38 bytes proc h x {expr$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. Commented May 28, 2017 at 20:30
• No problem! It looks better
– user63187
Commented May 28, 2017 at 21:30

05AB1E, 3 bytes

LzO


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1-indexed.

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


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Output is a fraction.

Vyxal, 3 bytes

ɾĖ∑


Lame answer, but range -> reciprocal -> sum -> implicit out, same as jelly

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Vyxals, 2 bytes

ɾĖ


2 Bytes w/ -s from @lyxal, -s sums the stack

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• 2 bytes Commented May 18, 2021 at 2:07

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


C, 54 bytes

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


Uses 1-indexed numbers.

f n=sum\$map(1/)[1..n]


Brachylog, 6 bytes

⟦₁/₁ᵐ+


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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


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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)
B  Return


Rust, 36 bytes

|n|(1..=n).map(|n|1./n as f64).sum()


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Input is an integer, if the input is <=0 the output is zero, and inputting one gets one.

J, 9 bytes

1#.1%1+i.


I'm only posting this because I think it looks really pretty.

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1#.1%1+i.
i.  NB. range 0..n-1
1%      NB. 1 divided by each element
1#.        NB. sum the result


Pyt, 3 bytes

ř⅟Ʃ
implicit input
ř       produces a list containing 1 to n
⅟      takes the reciprocals
Ʃ     sums the list
(implicit print)


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Scala, 47 bytes

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def h(n:Int):Double=if(n==0)0 else 1.0/n+h(n-1)


Thunno 2S, 2 bytes

RỊ


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Explanation

RỊ  # Implicit input
R   # One range
Ị  # Reciprocal
# Take the sum
# Implicit output


C (gcc), 35 bytes

float f(n){return n?1./n+f(--n):0;}


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