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
49.1k870156
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
$endgroup$
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
49.1k870156
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
$endgroup$
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
49.1k870156
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
$endgroup$
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
49.1k870156
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
edited Dec 8 '18 at 20:02
Blue
49.1k870156
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
edited Dec 8 '18 at 20:02
Blue
49.1k870156
edited Dec 8 '18 at 20:02
Blue
49.1k870156
edited Dec 8 '18 at 20:02
Blue
49.1k870156
49.1k870156
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
asked Dec 8 '18 at 19:55
TheoryQuest1TheoryQuest1
1498
1498
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031561%2fhow-do-we-compute-sum-n-1-infty-frac1ksn-for-k-an-integer-and%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031561%2fhow-do-we-compute-sum-n-1-infty-frac1ksn-for-k-an-integer-and%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Write $s=K^S$. Then $sum_{n=1}^{infty}s^{-n}$.
$endgroup$
– Dietrich Burde
Dec 8 '18 at 19:59