How do we compute $sum_{n=1}^{infty}frac{1}{K^{Sn}}$, for $K$ an integer and $S$ a complex number?
$begingroup$
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
edited Dec 8 '18 at 20:02
Blue
47.9k870153
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1266
$endgroup$
add a comment |
$begingroup$
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
edited Dec 8 '18 at 20:02
Blue
47.9k870153
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1266
$endgroup$
1
$begingroup$
Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 19:59
add a comment |
$begingroup$
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
edited Dec 8 '18 at 20:02
Blue
47.9k870153
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1266
$endgroup$
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
edited Dec 8 '18 at 20:02
Blue
47.9k870153
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1266
edited Dec 8 '18 at 20:02
Blue
47.9k870153
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1266
edited Dec 8 '18 at 20:02
Blue
47.9k870153
edited Dec 8 '18 at 20:02
Blue
47.9k870153
edited Dec 8 '18 at 20:02
Blue
47.9k870153
47.9k870153
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1266
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1266
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Looks like a geometric progression with quotient $1/K^S$.
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
$endgroup$
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
add a comment |
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1 Answer
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oldest
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1 Answer
1
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oldest
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$begingroup$
Looks like a geometric progression with quotient $1/K^S$.
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
$endgroup$
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
add a comment |
$begingroup$
Looks like a geometric progression with quotient $1/K^S$.
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
$endgroup$
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
add a comment |
$begingroup$
Looks like a geometric progression with quotient $1/K^S$.
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
$endgroup$
Looks like a geometric progression with quotient $1/K^S$.
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
answered Dec 8 '18 at 19:59
A. PongráczA. Pongrácz
5,9731929
5,9731929
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
add a comment |
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
add a comment |
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$begingroup$
Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 19:59