Using a For Loop to find a sequences expression [1/1+1/2+1/3…1/1000]

I am needing assistance for an assignment. I am needing to use python to figure out how to create a "for Loop" that will do the following:

Series 1/1+1/2+1/3+1/4+...+1/1000 (which is expressed as 1000. ∑ n=1. 1 n ≈. 7.49)

I am needing the program to loop through all of them, printing each number out.
Example:

998 7.483469855949342



999 7.48447086055343



1000 7.485470860550343

The basic what I currently got is

for x in range(1, 1000):

I don't know why but I just struggling to get this equation to work in my head. Any help would be greatly appreciated.

edited Nov 25 '18 at 8:51

lagom

3,142102038

asked Nov 25 '18 at 6:10

Anna Collins

488

add a comment |

I am needing assistance for an assignment. I am needing to use python to figure out how to create a "for Loop" that will do the following:

Series 1/1+1/2+1/3+1/4+...+1/1000 (which is expressed as 1000. ∑ n=1. 1 n ≈. 7.49)

I am needing the program to loop through all of them, printing each number out.
Example:

998 7.483469855949342



999 7.48447086055343



1000 7.485470860550343

The basic what I currently got is

for x in range(1, 1000):

I don't know why but I just struggling to get this equation to work in my head. Any help would be greatly appreciated.

edited Nov 25 '18 at 8:51

lagom

3,142102038

asked Nov 25 '18 at 6:10

Anna Collins

488

add a comment |

I am needing assistance for an assignment. I am needing to use python to figure out how to create a "for Loop" that will do the following:

Series 1/1+1/2+1/3+1/4+...+1/1000 (which is expressed as 1000. ∑ n=1. 1 n ≈. 7.49)

I am needing the program to loop through all of them, printing each number out.
Example:

998 7.483469855949342



999 7.48447086055343



1000 7.485470860550343

The basic what I currently got is

for x in range(1, 1000):

I don't know why but I just struggling to get this equation to work in my head. Any help would be greatly appreciated.

edited Nov 25 '18 at 8:51

lagom

3,142102038

asked Nov 25 '18 at 6:10

Anna Collins

488

I am needing assistance for an assignment. I am needing to use python to figure out how to create a "for Loop" that will do the following:

Series 1/1+1/2+1/3+1/4+...+1/1000 (which is expressed as 1000. ∑ n=1. 1 n ≈. 7.49)

I am needing the program to loop through all of them, printing each number out.
Example:

998 7.483469855949342



999 7.48447086055343



1000 7.485470860550343

The basic what I currently got is

for x in range(1, 1000):

I don't know why but I just struggling to get this equation to work in my head. Any help would be greatly appreciated.

python for-loop math sequence

edited Nov 25 '18 at 8:51

lagom

3,142102038

asked Nov 25 '18 at 6:10

Anna Collins

488

edited Nov 25 '18 at 8:51

lagom

3,142102038

asked Nov 25 '18 at 6:10

Anna Collins

488

edited Nov 25 '18 at 8:51

lagom

3,142102038

edited Nov 25 '18 at 8:51

lagom

3,142102038

edited Nov 25 '18 at 8:51

lagom

3,142102038

asked Nov 25 '18 at 6:10

Anna Collins

488

asked Nov 25 '18 at 6:10

Anna Collins

488

asked Nov 25 '18 at 6:10

Anna Collins

488

add a comment |

3 Answers
3

active

oldest

votes

You are going in the right direction. Before the for loop you would require a sum variable, where you would store the value of the summation of 1/x.

You can do that in a similar way:

sum = 0

for x in range(1, 1001):

    sum += (1/x)

    print(sum, x)

Here, I have initialized the sum variable to 0. After that, I iterate x over the values of [1, 1000] (both included). I find 1/x and add it to the sum. Next, I print the values, as you wanted.

NOTE: range(x, y) method gives you a range from x to y-1

answered Nov 25 '18 at 6:16

MaJoR

534115

Bravo! The only issue I see here is that your are storing values into memory. This means slowness if dealing with big data. ;)

– Prayson W. Daniel
Nov 25 '18 at 6:49

1

@PraysonW.Daniel, you are right, however this seems to be beginner's question. We should not impose the effects of sluggishness of storing variables in memory right now (at this level), I guess.

– MaJoR
Nov 25 '18 at 6:54

1

Agree. I think for tutorials purposes, it is okay but were it to be a production code, efficiency is to be preferred.

– Prayson W. Daniel
Nov 25 '18 at 6:57

add a comment |

itertools are your best friend. The proposed answers is correct but would be slow for big data. If I were you I would do:

import itertools



a = map(lambda x:1/x,range(1,1001))



#print(list(itertools.accumulate(a)))



for i, j in enumerate(1,itertools.accumulate(a)):

    print(i,j)

Explaination: lambda x:1/x creates on-fly function that would transform n to 1/n. map maps that function to the range of value starting from 1 to 1000. I then pass this to accumulating 1/1+1/2..... ;)

edited Nov 25 '18 at 6:47

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

add a comment |

Keep in mind that python2 will return 0 for 1/x and your sum will lead to 1 at the end. For getting float output(i.e. 0.25 for 1/4) one of the numbers have to be converted to float(either 1 or either x). Hence, the right way would be

sum = 0

for x in range(1, 1001):

    sum += (float(1)/x)

    print(sum, x)

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

Or you can import division from the future. from __future__ import division. The Python 2 division would be replaced by 3 and business as usual :)

– Prayson W. Daniel
Nov 25 '18 at 6:53

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You are going in the right direction. Before the for loop you would require a sum variable, where you would store the value of the summation of 1/x.

You can do that in a similar way:

sum = 0

for x in range(1, 1001):

    sum += (1/x)

    print(sum, x)

Here, I have initialized the sum variable to 0. After that, I iterate x over the values of [1, 1000] (both included). I find 1/x and add it to the sum. Next, I print the values, as you wanted.

NOTE: range(x, y) method gives you a range from x to y-1

answered Nov 25 '18 at 6:16

MaJoR

534115

Bravo! The only issue I see here is that your are storing values into memory. This means slowness if dealing with big data. ;)

– Prayson W. Daniel
Nov 25 '18 at 6:49

1

@PraysonW.Daniel, you are right, however this seems to be beginner's question. We should not impose the effects of sluggishness of storing variables in memory right now (at this level), I guess.

– MaJoR
Nov 25 '18 at 6:54

1

Agree. I think for tutorials purposes, it is okay but were it to be a production code, efficiency is to be preferred.

– Prayson W. Daniel
Nov 25 '18 at 6:57

add a comment |

You are going in the right direction. Before the for loop you would require a sum variable, where you would store the value of the summation of 1/x.

You can do that in a similar way:

sum = 0

for x in range(1, 1001):

    sum += (1/x)

    print(sum, x)

Here, I have initialized the sum variable to 0. After that, I iterate x over the values of [1, 1000] (both included). I find 1/x and add it to the sum. Next, I print the values, as you wanted.

NOTE: range(x, y) method gives you a range from x to y-1

answered Nov 25 '18 at 6:16

MaJoR

534115

Bravo! The only issue I see here is that your are storing values into memory. This means slowness if dealing with big data. ;)

– Prayson W. Daniel
Nov 25 '18 at 6:49

1

@PraysonW.Daniel, you are right, however this seems to be beginner's question. We should not impose the effects of sluggishness of storing variables in memory right now (at this level), I guess.

– MaJoR
Nov 25 '18 at 6:54

1

Agree. I think for tutorials purposes, it is okay but were it to be a production code, efficiency is to be preferred.

– Prayson W. Daniel
Nov 25 '18 at 6:57

add a comment |

You are going in the right direction. Before the for loop you would require a sum variable, where you would store the value of the summation of 1/x.

You can do that in a similar way:

sum = 0

for x in range(1, 1001):

    sum += (1/x)

    print(sum, x)

Here, I have initialized the sum variable to 0. After that, I iterate x over the values of [1, 1000] (both included). I find 1/x and add it to the sum. Next, I print the values, as you wanted.

NOTE: range(x, y) method gives you a range from x to y-1

answered Nov 25 '18 at 6:16

MaJoR

534115

You are going in the right direction. Before the for loop you would require a sum variable, where you would store the value of the summation of 1/x.

You can do that in a similar way:

sum = 0

for x in range(1, 1001):

    sum += (1/x)

    print(sum, x)

Here, I have initialized the sum variable to 0. After that, I iterate x over the values of [1, 1000] (both included). I find 1/x and add it to the sum. Next, I print the values, as you wanted.

NOTE: range(x, y) method gives you a range from x to y-1

answered Nov 25 '18 at 6:16

MaJoR

534115

answered Nov 25 '18 at 6:16

MaJoR

534115

answered Nov 25 '18 at 6:16

MaJoR

534115

answered Nov 25 '18 at 6:16

MaJoR

534115

Bravo! The only issue I see here is that your are storing values into memory. This means slowness if dealing with big data. ;)

– Prayson W. Daniel
Nov 25 '18 at 6:49

1

@PraysonW.Daniel, you are right, however this seems to be beginner's question. We should not impose the effects of sluggishness of storing variables in memory right now (at this level), I guess.

– MaJoR
Nov 25 '18 at 6:54

1

Agree. I think for tutorials purposes, it is okay but were it to be a production code, efficiency is to be preferred.

– Prayson W. Daniel
Nov 25 '18 at 6:57

add a comment |

Bravo! The only issue I see here is that your are storing values into memory. This means slowness if dealing with big data. ;)

– Prayson W. Daniel
Nov 25 '18 at 6:49

1

@PraysonW.Daniel, you are right, however this seems to be beginner's question. We should not impose the effects of sluggishness of storing variables in memory right now (at this level), I guess.

– MaJoR
Nov 25 '18 at 6:54

1

Agree. I think for tutorials purposes, it is okay but were it to be a production code, efficiency is to be preferred.

– Prayson W. Daniel
Nov 25 '18 at 6:57

Bravo! The only issue I see here is that your are storing values into memory. This means slowness if dealing with big data. ;)

– Prayson W. Daniel
Nov 25 '18 at 6:49

@PraysonW.Daniel, you are right, however this seems to be beginner's question. We should not impose the effects of sluggishness of storing variables in memory right now (at this level), I guess.

– MaJoR
Nov 25 '18 at 6:54

Agree. I think for tutorials purposes, it is okay but were it to be a production code, efficiency is to be preferred.

– Prayson W. Daniel
Nov 25 '18 at 6:57

add a comment |

itertools are your best friend. The proposed answers is correct but would be slow for big data. If I were you I would do:

import itertools



a = map(lambda x:1/x,range(1,1001))



#print(list(itertools.accumulate(a)))



for i, j in enumerate(1,itertools.accumulate(a)):

    print(i,j)

edited Nov 25 '18 at 6:47

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

add a comment |

itertools are your best friend. The proposed answers is correct but would be slow for big data. If I were you I would do:

import itertools



a = map(lambda x:1/x,range(1,1001))



#print(list(itertools.accumulate(a)))



for i, j in enumerate(1,itertools.accumulate(a)):

    print(i,j)

edited Nov 25 '18 at 6:47

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

add a comment |

itertools are your best friend. The proposed answers is correct but would be slow for big data. If I were you I would do:

import itertools



a = map(lambda x:1/x,range(1,1001))



#print(list(itertools.accumulate(a)))



for i, j in enumerate(1,itertools.accumulate(a)):

    print(i,j)

edited Nov 25 '18 at 6:47

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

itertools are your best friend. The proposed answers is correct but would be slow for big data. If I were you I would do:

import itertools



a = map(lambda x:1/x,range(1,1001))



#print(list(itertools.accumulate(a)))



for i, j in enumerate(1,itertools.accumulate(a)):

    print(i,j)

edited Nov 25 '18 at 6:47

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

edited Nov 25 '18 at 6:47

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

answered Nov 25 '18 at 6:41

Prayson W. Daniel

2,10111319

add a comment |

sum = 0

for x in range(1, 1001):

    sum += (float(1)/x)

    print(sum, x)

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

Or you can import division from the future. from __future__ import division. The Python 2 division would be replaced by 3 and business as usual :)

– Prayson W. Daniel
Nov 25 '18 at 6:53

add a comment |

sum = 0

for x in range(1, 1001):

    sum += (float(1)/x)

    print(sum, x)

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

Or you can import division from the future. from __future__ import division. The Python 2 division would be replaced by 3 and business as usual :)

– Prayson W. Daniel
Nov 25 '18 at 6:53

add a comment |

sum = 0

for x in range(1, 1001):

    sum += (float(1)/x)

    print(sum, x)

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

sum = 0

for x in range(1, 1001):

    sum += (float(1)/x)

    print(sum, x)

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

answered Nov 25 '18 at 6:25

Biswadip Mandal

1809

Or you can import division from the future. from __future__ import division. The Python 2 division would be replaced by 3 and business as usual :)

– Prayson W. Daniel
Nov 25 '18 at 6:53

add a comment |

Or you can import division from the future. from __future__ import division. The Python 2 division would be replaced by 3 and business as usual :)

– Prayson W. Daniel
Nov 25 '18 at 6:53

Or you can import division from the future. from __future__ import division. The Python 2 division would be replaced by 3 and business as usual :)

– Prayson W. Daniel
Nov 25 '18 at 6:53

add a comment |

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