Determining whether $sum_{n=2}^inftyfrac{sin(npi/12)}{ln n}$ converges
I'm having issues with determining convergence/divergence of alternating series that use sine and cosine. I'm perfectly clear of how to handle ones with $(-1)^{n+1}$ (and similar) by performing the Absolute Convergence Test and by applying Leibnitz's theorem, but sine and cosine ones are a totally different story.
I simply don't know where to start on this one, for example.
$$sum_{n=2}^inftyfrac{sindfrac{npi}{12}}{ln n}$$
sequences-and-series convergence
|
show 3 more comments
I'm having issues with determining convergence/divergence of alternating series that use sine and cosine. I'm perfectly clear of how to handle ones with $(-1)^{n+1}$ (and similar) by performing the Absolute Convergence Test and by applying Leibnitz's theorem, but sine and cosine ones are a totally different story.
I simply don't know where to start on this one, for example.
$$sum_{n=2}^inftyfrac{sindfrac{npi}{12}}{ln n}$$
sequences-and-series convergence
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 25 at 15:00
Thanks for the welcome! Yeah, I'll take your advice definitely, just need some more time to get to know things.
– Rose
Nov 25 at 15:02
I've used MathJax to display the series. Feel free to make the rest of the post less wordy, or to clarify whether there's anything you want evaluated besides that one series.
– J.G.
Nov 25 at 15:03
Thank You! No, that'd be pretty much it I'm interested in right now.
– Rose
Nov 25 at 15:06
Observe $displaystylesin(npi/12)$ is periodic so its sum cannot be greater than $sum_{n=1}^{12} sin(npi/{12})$.
– Yadati Kiran
Nov 25 at 15:15
|
show 3 more comments
I'm having issues with determining convergence/divergence of alternating series that use sine and cosine. I'm perfectly clear of how to handle ones with $(-1)^{n+1}$ (and similar) by performing the Absolute Convergence Test and by applying Leibnitz's theorem, but sine and cosine ones are a totally different story.
I simply don't know where to start on this one, for example.
$$sum_{n=2}^inftyfrac{sindfrac{npi}{12}}{ln n}$$
sequences-and-series convergence
I'm having issues with determining convergence/divergence of alternating series that use sine and cosine. I'm perfectly clear of how to handle ones with $(-1)^{n+1}$ (and similar) by performing the Absolute Convergence Test and by applying Leibnitz's theorem, but sine and cosine ones are a totally different story.
I simply don't know where to start on this one, for example.
$$sum_{n=2}^inftyfrac{sindfrac{npi}{12}}{ln n}$$
sequences-and-series convergence
sequences-and-series convergence
edited Nov 25 at 15:10
Blue
47.5k870150
47.5k870150
asked Nov 25 at 14:58
Rose
65
65
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 25 at 15:00
Thanks for the welcome! Yeah, I'll take your advice definitely, just need some more time to get to know things.
– Rose
Nov 25 at 15:02
I've used MathJax to display the series. Feel free to make the rest of the post less wordy, or to clarify whether there's anything you want evaluated besides that one series.
– J.G.
Nov 25 at 15:03
Thank You! No, that'd be pretty much it I'm interested in right now.
– Rose
Nov 25 at 15:06
Observe $displaystylesin(npi/12)$ is periodic so its sum cannot be greater than $sum_{n=1}^{12} sin(npi/{12})$.
– Yadati Kiran
Nov 25 at 15:15
|
show 3 more comments
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 25 at 15:00
Thanks for the welcome! Yeah, I'll take your advice definitely, just need some more time to get to know things.
– Rose
Nov 25 at 15:02
I've used MathJax to display the series. Feel free to make the rest of the post less wordy, or to clarify whether there's anything you want evaluated besides that one series.
– J.G.
Nov 25 at 15:03
Thank You! No, that'd be pretty much it I'm interested in right now.
– Rose
Nov 25 at 15:06
Observe $displaystylesin(npi/12)$ is periodic so its sum cannot be greater than $sum_{n=1}^{12} sin(npi/{12})$.
– Yadati Kiran
Nov 25 at 15:15
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 25 at 15:00
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 25 at 15:00
Thanks for the welcome! Yeah, I'll take your advice definitely, just need some more time to get to know things.
– Rose
Nov 25 at 15:02
Thanks for the welcome! Yeah, I'll take your advice definitely, just need some more time to get to know things.
– Rose
Nov 25 at 15:02
I've used MathJax to display the series. Feel free to make the rest of the post less wordy, or to clarify whether there's anything you want evaluated besides that one series.
– J.G.
Nov 25 at 15:03
I've used MathJax to display the series. Feel free to make the rest of the post less wordy, or to clarify whether there's anything you want evaluated besides that one series.
– J.G.
Nov 25 at 15:03
Thank You! No, that'd be pretty much it I'm interested in right now.
– Rose
Nov 25 at 15:06
Thank You! No, that'd be pretty much it I'm interested in right now.
– Rose
Nov 25 at 15:06
Observe $displaystylesin(npi/12)$ is periodic so its sum cannot be greater than $sum_{n=1}^{12} sin(npi/{12})$.
– Yadati Kiran
Nov 25 at 15:15
Observe $displaystylesin(npi/12)$ is periodic so its sum cannot be greater than $sum_{n=1}^{12} sin(npi/{12})$.
– Yadati Kiran
Nov 25 at 15:15
|
show 3 more comments
1 Answer
1
active
oldest
votes
We use the Dirichlet's test.
For that, we observe that $frac{1}{log (n)}$ is decreasing and tending to $0$ as $nto infty.$ Next we have to show that
$$S_M=sum_{n=1}^{M}sin(npi/12)$$ is bounded.
For this use the following equation:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
Ah, Dirichlet's theorem, I'd never think of it, thanks! Could you just explain the way you prove that the partial sums are bounded?
– Rose
Nov 25 at 15:41
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%2f3012942%2fdetermining-whether-sum-n-2-infty-frac-sinn-pi-12-ln-n-converges%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
We use the Dirichlet's test.
For that, we observe that $frac{1}{log (n)}$ is decreasing and tending to $0$ as $nto infty.$ Next we have to show that
$$S_M=sum_{n=1}^{M}sin(npi/12)$$ is bounded.
For this use the following equation:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
Ah, Dirichlet's theorem, I'd never think of it, thanks! Could you just explain the way you prove that the partial sums are bounded?
– Rose
Nov 25 at 15:41
add a comment |
We use the Dirichlet's test.
For that, we observe that $frac{1}{log (n)}$ is decreasing and tending to $0$ as $nto infty.$ Next we have to show that
$$S_M=sum_{n=1}^{M}sin(npi/12)$$ is bounded.
For this use the following equation:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
Ah, Dirichlet's theorem, I'd never think of it, thanks! Could you just explain the way you prove that the partial sums are bounded?
– Rose
Nov 25 at 15:41
add a comment |
We use the Dirichlet's test.
For that, we observe that $frac{1}{log (n)}$ is decreasing and tending to $0$ as $nto infty.$ Next we have to show that
$$S_M=sum_{n=1}^{M}sin(npi/12)$$ is bounded.
For this use the following equation:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
We use the Dirichlet's test.
For that, we observe that $frac{1}{log (n)}$ is decreasing and tending to $0$ as $nto infty.$ Next we have to show that
$$S_M=sum_{n=1}^{M}sin(npi/12)$$ is bounded.
For this use the following equation:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
answered Nov 25 at 15:09
Hello_World
3,83521630
3,83521630
Ah, Dirichlet's theorem, I'd never think of it, thanks! Could you just explain the way you prove that the partial sums are bounded?
– Rose
Nov 25 at 15:41
add a comment |
Ah, Dirichlet's theorem, I'd never think of it, thanks! Could you just explain the way you prove that the partial sums are bounded?
– Rose
Nov 25 at 15:41
Ah, Dirichlet's theorem, I'd never think of it, thanks! Could you just explain the way you prove that the partial sums are bounded?
– Rose
Nov 25 at 15:41
Ah, Dirichlet's theorem, I'd never think of it, thanks! Could you just explain the way you prove that the partial sums are bounded?
– Rose
Nov 25 at 15:41
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3012942%2fdetermining-whether-sum-n-2-infty-frac-sinn-pi-12-ln-n-converges%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
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 25 at 15:00
Thanks for the welcome! Yeah, I'll take your advice definitely, just need some more time to get to know things.
– Rose
Nov 25 at 15:02
I've used MathJax to display the series. Feel free to make the rest of the post less wordy, or to clarify whether there's anything you want evaluated besides that one series.
– J.G.
Nov 25 at 15:03
Thank You! No, that'd be pretty much it I'm interested in right now.
– Rose
Nov 25 at 15:06
Observe $displaystylesin(npi/12)$ is periodic so its sum cannot be greater than $sum_{n=1}^{12} sin(npi/{12})$.
– Yadati Kiran
Nov 25 at 15:15