Partial Sum of Harmonic Sequence!

Question

Definition

In Mathematics, Harmonic Sequence refers to a sequence where

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

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.

---

Additional Rules

• The input number can be either of 0-indexed or 1-indexed. You must specify that in your post.

• You must not use a built-in to calculate the partial sum of the first n elements. (Yeah, it's for you Mathematica!)

• You must not abuse native number types to trivialize the problem.

• Standard Loopholes apply.

---

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 code-golf, so the shortest code in bytes wins!

Answer 1

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

Answer 2

Python 3, 27 bytes

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

Answer 3

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))

Answer 4

Mathematica, 21 20 16 bytes

This solution is 1-indexed.

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

Answer 5

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

Answer 6

Pari/GP, 18 bytes

n->sum(i=1,n,1./i)

1-indexing.

Try it online!

Answer 7

CJam, 11 bytes

1.ri,:)f/:+

Try it online!

1-indexed.

Answer 8

PHP, 33 Bytes

1-indexing

for(;$i++<$argn;)$s+=1/$i;echo$s;

Try it online!

Answer 9

Japt -x, 8 6 5 3 bytes

õpJ

With some thanks to ETHproductions

Try it online

Answer 10

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

Answer 11

05AB1E, 3 bytes

LzO

Try it online!

1-indexed.

Answer 12

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.

Answer 13

R, 15 bytes

sum(1/1:scan())

Try it online!

Answer 14

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.

Answer 15

Pyth, 5 bytes

scL1S

Try it here.

1-indexed.

Answer 16

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

Answer 17

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

Answer 18

C, 54 bytes

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

Uses 1-indexed numbers.

Answer 19

Brachylog, 6 bytes

⟦₁/₁ᵐ+

Try it online!

This is 1-indexed.

Explanation

⟦₁         Range [1, …, Input]
    ᵐ      Map:
  /₁         Inverse
     +     Sum

Answer 20

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

Answer 21

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

Answer 22

Tcl, 61 bytes

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

Try it online!

Answer 23

Haskell, 21 bytes

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

Answer 24

C (gcc), 35 bytes

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

Try it online!

Answer 25

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.