How do we compute $sum_{n=1}^{infty}frac{1}{K^{Sn}}$, for $K$ an integer and $S$ a complex number?












-2












$begingroup$


Consider the series:
$$sum_{n=1}^{infty}frac{1}{K^{Sn}}$$



where $K$ is some integer constant and $S$ is a complex number.



How we evaluate (compute sum of) this series where S is a complex number. Can we use integration? (I am unsure how we will do that as S is a complex number.)



Can someone please help with the formula? I am new to complex analysis thus struggling.






sequences-and-series complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 19:59
















-2












$begingroup$


Consider the series:
$$sum_{n=1}^{infty}frac{1}{K^{Sn}}$$



where $K$ is some integer constant and $S$ is a complex number.



How we evaluate (compute sum of) this series where S is a complex number. Can we use integration? (I am unsure how we will do that as S is a complex number.)



Can someone please help with the formula? I am new to complex analysis thus struggling.






sequences-and-series complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 19:59














-2












-2








-2





$begingroup$


Consider the series:
$$sum_{n=1}^{infty}frac{1}{K^{Sn}}$$



where $K$ is some integer constant and $S$ is a complex number.



How we evaluate (compute sum of) this series where S is a complex number. Can we use integration? (I am unsure how we will do that as S is a complex number.)



Can someone please help with the formula? I am new to complex analysis thus struggling.






sequences-and-series complex-analysis







share|cite|improve this question











$endgroup$




Consider the series:
$$sum_{n=1}^{infty}frac{1}{K^{Sn}}$$



where $K$ is some integer constant and $S$ is a complex number.



How we evaluate (compute sum of) this series where S is a complex number. Can we use integration? (I am unsure how we will do that as S is a complex number.)



Can someone please help with the formula? I am new to complex analysis thus struggling.






sequences-and-series complex-analysis




sequences-and-series complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 20:02









Blue

48k870153




48k870153










asked Dec 8 '18 at 19:55









TheoryQuest1TheoryQuest1

1266




1266








  • 1




    $begingroup$
    Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 19:59














  • 1




    $begingroup$
    Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
    $endgroup$
    – Dietrich Burde
    Dec 8 '18 at 19:59








1




1




$begingroup$
Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 19:59




$begingroup$
Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 19:59










1 Answer
1






active

oldest

votes


















3












$begingroup$

Looks like a geometric progression with quotient $1/K^S$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Indeed it is. So, can we simply evaluate the sum as: $1/(1-r)$ where $r=1/K^S$ ?
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:10










  • $begingroup$
    That is correct.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 20:12










  • $begingroup$
    Thanks a lot...
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:12











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Looks like a geometric progression with quotient $1/K^S$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Indeed it is. So, can we simply evaluate the sum as: $1/(1-r)$ where $r=1/K^S$ ?
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:10










  • $begingroup$
    That is correct.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 20:12










  • $begingroup$
    Thanks a lot...
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:12
















3












$begingroup$

Looks like a geometric progression with quotient $1/K^S$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Indeed it is. So, can we simply evaluate the sum as: $1/(1-r)$ where $r=1/K^S$ ?
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:10










  • $begingroup$
    That is correct.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 20:12










  • $begingroup$
    Thanks a lot...
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:12














3












3








3





$begingroup$

Looks like a geometric progression with quotient $1/K^S$.






share|cite|improve this answer









$endgroup$



Looks like a geometric progression with quotient $1/K^S$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 19:59









A. PongráczA. Pongrácz

5,9631929




5,9631929








  • 1




    $begingroup$
    Indeed it is. So, can we simply evaluate the sum as: $1/(1-r)$ where $r=1/K^S$ ?
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:10










  • $begingroup$
    That is correct.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 20:12










  • $begingroup$
    Thanks a lot...
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:12














  • 1




    $begingroup$
    Indeed it is. So, can we simply evaluate the sum as: $1/(1-r)$ where $r=1/K^S$ ?
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:10










  • $begingroup$
    That is correct.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 20:12










  • $begingroup$
    Thanks a lot...
    $endgroup$
    – TheoryQuest1
    Dec 8 '18 at 20:12








1




1




$begingroup$
Indeed it is. So, can we simply evaluate the sum as: $1/(1-r)$ where $r=1/K^S$ ?
$endgroup$
– TheoryQuest1
Dec 8 '18 at 20:10




$begingroup$
Indeed it is. So, can we simply evaluate the sum as: $1/(1-r)$ where $r=1/K^S$ ?
$endgroup$
– TheoryQuest1
Dec 8 '18 at 20:10












$begingroup$
That is correct.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:12




$begingroup$
That is correct.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:12












$begingroup$
Thanks a lot...
$endgroup$
– TheoryQuest1
Dec 8 '18 at 20:12




$begingroup$
Thanks a lot...
$endgroup$
– TheoryQuest1
Dec 8 '18 at 20:12


















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