How to show that $sum_{k=1}^ne^{ik}=frac{(e^{in}-1)e^i}{e^i-1}$












0












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The question fits in the title: what tool(s) is/are needed to deduce this identity?
$$sum_{k=1}^ne^{ik}=frac{(e^{in}-1)e^i}{e^i-1}.$$
This is the last step in my solution to another problem; I've tried playing around with the Binomial theorem to no avail and don't see what to try next. A good hint will suffice!






complex-numbers







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  • 1




    $begingroup$
    Keyword: geometric series. For every $rne1$, $$sum_{k=1}^nr^k=rfrac{1-r^n}{1-r}$$
    $endgroup$
    – Did
    Jan 27 at 11:24










  • $begingroup$
    finite geometric series
    $endgroup$
    – Wesley Strik
    Jan 27 at 11:27
















0












$begingroup$


The question fits in the title: what tool(s) is/are needed to deduce this identity?
$$sum_{k=1}^ne^{ik}=frac{(e^{in}-1)e^i}{e^i-1}.$$
This is the last step in my solution to another problem; I've tried playing around with the Binomial theorem to no avail and don't see what to try next. A good hint will suffice!






complex-numbers







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Keyword: geometric series. For every $rne1$, $$sum_{k=1}^nr^k=rfrac{1-r^n}{1-r}$$
    $endgroup$
    – Did
    Jan 27 at 11:24










  • $begingroup$
    finite geometric series
    $endgroup$
    – Wesley Strik
    Jan 27 at 11:27














0












0








0





$begingroup$


The question fits in the title: what tool(s) is/are needed to deduce this identity?
$$sum_{k=1}^ne^{ik}=frac{(e^{in}-1)e^i}{e^i-1}.$$
This is the last step in my solution to another problem; I've tried playing around with the Binomial theorem to no avail and don't see what to try next. A good hint will suffice!






complex-numbers







share|cite|improve this question









$endgroup$




The question fits in the title: what tool(s) is/are needed to deduce this identity?
$$sum_{k=1}^ne^{ik}=frac{(e^{in}-1)e^i}{e^i-1}.$$
This is the last step in my solution to another problem; I've tried playing around with the Binomial theorem to no avail and don't see what to try next. A good hint will suffice!






complex-numbers




complex-numbers






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share|cite|improve this question











share|cite|improve this question




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asked Jan 27 at 11:19









Sup BroSup Bro

396214




396214








  • 1




    $begingroup$
    Keyword: geometric series. For every $rne1$, $$sum_{k=1}^nr^k=rfrac{1-r^n}{1-r}$$
    $endgroup$
    – Did
    Jan 27 at 11:24










  • $begingroup$
    finite geometric series
    $endgroup$
    – Wesley Strik
    Jan 27 at 11:27














  • 1




    $begingroup$
    Keyword: geometric series. For every $rne1$, $$sum_{k=1}^nr^k=rfrac{1-r^n}{1-r}$$
    $endgroup$
    – Did
    Jan 27 at 11:24










  • $begingroup$
    finite geometric series
    $endgroup$
    – Wesley Strik
    Jan 27 at 11:27








1




1




$begingroup$
Keyword: geometric series. For every $rne1$, $$sum_{k=1}^nr^k=rfrac{1-r^n}{1-r}$$
$endgroup$
– Did
Jan 27 at 11:24




$begingroup$
Keyword: geometric series. For every $rne1$, $$sum_{k=1}^nr^k=rfrac{1-r^n}{1-r}$$
$endgroup$
– Did
Jan 27 at 11:24












$begingroup$
finite geometric series
$endgroup$
– Wesley Strik
Jan 27 at 11:27




$begingroup$
finite geometric series
$endgroup$
– Wesley Strik
Jan 27 at 11:27










2 Answers
2






active

oldest

votes


















3












$begingroup$

This is just geometric series with $a = e^i$ and $r = e^i$.



Apply the summation formula for a finite geometric series.



Hint: $sum_{k=0}^{n} ar^k = displaystylefrac{a(r^n-1)}{r-1} text{ for }r>1$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Perfect! I feel stupid now, I blame the lack of sleep. I shall accept as solved when MSE lets me
    $endgroup$
    – Sup Bro
    Jan 27 at 11:25



















2












$begingroup$

Hint:



$sum_{k=1}^ne^{ik}=e^i-e^{i(n+1)}+e^icdotsum_{k=1}^ne^{ik}$



Note:
This is a particular case of the geometric summation ($sum a^i$). The same reasoning exposed for $e^i$ is valid for a general $a$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    This is just geometric series with $a = e^i$ and $r = e^i$.



    Apply the summation formula for a finite geometric series.



    Hint: $sum_{k=0}^{n} ar^k = displaystylefrac{a(r^n-1)}{r-1} text{ for }r>1$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Perfect! I feel stupid now, I blame the lack of sleep. I shall accept as solved when MSE lets me
      $endgroup$
      – Sup Bro
      Jan 27 at 11:25
















    3












    $begingroup$

    This is just geometric series with $a = e^i$ and $r = e^i$.



    Apply the summation formula for a finite geometric series.



    Hint: $sum_{k=0}^{n} ar^k = displaystylefrac{a(r^n-1)}{r-1} text{ for }r>1$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Perfect! I feel stupid now, I blame the lack of sleep. I shall accept as solved when MSE lets me
      $endgroup$
      – Sup Bro
      Jan 27 at 11:25














    3












    3








    3





    $begingroup$

    This is just geometric series with $a = e^i$ and $r = e^i$.



    Apply the summation formula for a finite geometric series.



    Hint: $sum_{k=0}^{n} ar^k = displaystylefrac{a(r^n-1)}{r-1} text{ for }r>1$






    share|cite|improve this answer











    $endgroup$



    This is just geometric series with $a = e^i$ and $r = e^i$.



    Apply the summation formula for a finite geometric series.



    Hint: $sum_{k=0}^{n} ar^k = displaystylefrac{a(r^n-1)}{r-1} text{ for }r>1$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 27 at 11:41

























    answered Jan 27 at 11:24









    Exp ikxExp ikx

    4489




    4489








    • 1




      $begingroup$
      Perfect! I feel stupid now, I blame the lack of sleep. I shall accept as solved when MSE lets me
      $endgroup$
      – Sup Bro
      Jan 27 at 11:25














    • 1




      $begingroup$
      Perfect! I feel stupid now, I blame the lack of sleep. I shall accept as solved when MSE lets me
      $endgroup$
      – Sup Bro
      Jan 27 at 11:25








    1




    1




    $begingroup$
    Perfect! I feel stupid now, I blame the lack of sleep. I shall accept as solved when MSE lets me
    $endgroup$
    – Sup Bro
    Jan 27 at 11:25




    $begingroup$
    Perfect! I feel stupid now, I blame the lack of sleep. I shall accept as solved when MSE lets me
    $endgroup$
    – Sup Bro
    Jan 27 at 11:25











    2












    $begingroup$

    Hint:



    $sum_{k=1}^ne^{ik}=e^i-e^{i(n+1)}+e^icdotsum_{k=1}^ne^{ik}$



    Note:
    This is a particular case of the geometric summation ($sum a^i$). The same reasoning exposed for $e^i$ is valid for a general $a$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint:



      $sum_{k=1}^ne^{ik}=e^i-e^{i(n+1)}+e^icdotsum_{k=1}^ne^{ik}$



      Note:
      This is a particular case of the geometric summation ($sum a^i$). The same reasoning exposed for $e^i$ is valid for a general $a$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint:



        $sum_{k=1}^ne^{ik}=e^i-e^{i(n+1)}+e^icdotsum_{k=1}^ne^{ik}$



        Note:
        This is a particular case of the geometric summation ($sum a^i$). The same reasoning exposed for $e^i$ is valid for a general $a$






        share|cite|improve this answer









        $endgroup$



        Hint:



        $sum_{k=1}^ne^{ik}=e^i-e^{i(n+1)}+e^icdotsum_{k=1}^ne^{ik}$



        Note:
        This is a particular case of the geometric summation ($sum a^i$). The same reasoning exposed for $e^i$ is valid for a general $a$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 27 at 11:22









        Gabriele CasseseGabriele Cassese

        1,171316




        1,171316






























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