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












1















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







share|improve this question





























    1















    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







    share|improve this question



























      1












      1








      1








      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







      share|improve this question
















      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




      python for-loop math sequence






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 25 '18 at 8:51









      lagom

      3,142102038




      3,142102038










      asked Nov 25 '18 at 6:10









      Anna CollinsAnna Collins

      488




      488
























          3 Answers
          3






          active

          oldest

          votes


















          4














          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






          share|improve this answer
























          • 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



















          1














          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..... ;)






          share|improve this answer

































            0














            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)





            share|improve this answer
























            • 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













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            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






            share|improve this answer
























            • 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
















            4














            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






            share|improve this answer
























            • 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














            4












            4








            4







            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






            share|improve this answer













            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







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 25 '18 at 6:16









            MaJoRMaJoR

            534115




            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



















            • 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





            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




            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





            @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




            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





            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













            1














            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..... ;)






            share|improve this answer






























              1














              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..... ;)






              share|improve this answer




























                1












                1








                1







                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..... ;)






                share|improve this answer















                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..... ;)







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 25 '18 at 6:47

























                answered Nov 25 '18 at 6:41









                Prayson W. DanielPrayson W. Daniel

                2,10111319




                2,10111319























                    0














                    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)





                    share|improve this answer
























                    • 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


















                    0














                    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)





                    share|improve this answer
























                    • 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
















                    0












                    0








                    0







                    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)





                    share|improve this answer













                    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)






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 25 '18 at 6:25









                    Biswadip MandalBiswadip Mandal

                    1809




                    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





















                    • 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







                    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




















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