The most unreliable gun ever [duplicate]

Question

You are given a very special gun with a full magazine.

Let n be the initial number of bullets in the magazine and i the number of bullets left.

That gun is really unreliable, hence each time you shoot, you have a i/n chance to successfully shoot. The fewer bullets you have left, the more tries it requires to shoot.

The goal is to find the average number of attempts to shoot before running out of ammo.

Example

You start with 3 bullets (n=3). Your first shot is always successful. You have now 2 bullets left. You will shoot first with a probability of 2/3 and misfire with 1/3. The probability of emptying your magazine in just 3 tries (no misfires) is (3/3) * (2/3) * (1/3).

The average number of tries before emptying your magazine for this example is 5.5.

Test Cases

f(2) = 3.000
f(3) = 5.500
f(4) = 8.330
f(10) ~= 29.290

Answer 1

05AB1E, 3 bytes

L/O

Try it online! or as a Test suite

Explanation

L     # range [1 ... n]
 /    # divide n by each in above list
  O   # sum

Answer 2

Haskell, 22 21 bytes

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

Try it online (TIO)!

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Previous, more self-explanatory version:

f n=sum[n/i|i<-[1..n]]

Answer 3

Ohm, 4 bytes

D@/Σ

Try it online!

Inspired by Emigna's answer

Answer 4

Japt, 6 4+1= 5 bytes

Inspired by Emigna's 05AB1E solution.

+1 byte for the -x flag.

õ@/X

Try it online

Answer 5

PHP, 37 Bytes

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

Try it online!

Answer 6

Aceto, 41 34 bytes

Based on Emigna's insight

Could almost definitely be golfed more.

  9jpX
 {:s_=1(
 &_L+&l@
iM!l
rLz@

Rough explanation

Part 1: Making a range and dividing by each of the numbers:

 {:s
 &_L
iM!l
rLz@

Part 2: Summing them up

    _=1(
    +&l@

Part 3: Printing, exiting, and making sure we don't visit this earlier

  9jpX

Answer 7

CJam, 11 bytes

rid_,:)f/:+

Try it online!

Inspired by Emigna's answer

Answer 8

jq, 30 characters

. as$n|[range(1;.+1)|$n/.]|add

Sample run:

bash-4.4$ jq '. as$n|[range(1;.+1)|$n/.]|add' <<< 10
29.289682539682538

Try in jq‣play

Answer 9

JavaScript (ES6), 44 34 27 26 bytes

Also inspired by Emigna's 05AB1E solution.

f=(n,i=n)=>i&&n/i+f(n,--i)

_{(Look at me , going for recursion over array mapping for once!)}

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Try it

f=(n,i=n)=>i&&n/i+f(n,--i)
oninput=_=>o.innerText=f(+i.value)
o.innerText=f(i.value=1)

<input id=i type=number><pre id=o>