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
edited Nov 25 at 15:10
Blue
47.5k870151
asked Nov 25 at 14:58
Rose
65
|
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
edited Nov 25 at 15:10
Blue
47.5k870151
asked Nov 25 at 14:58
Rose
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
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
edited Nov 25 at 15:10
Blue
47.5k870151
asked Nov 25 at 14:58
Rose
65
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
47.5k870151
asked Nov 25 at 14:58
Rose
65
edited Nov 25 at 15:10
Blue
47.5k870151
asked Nov 25 at 14:58
Rose
65
edited Nov 25 at 15:10
Blue
47.5k870151
edited Nov 25 at 15:10
Blue
47.5k870151
edited Nov 25 at 15:10
Blue
47.5k870151
47.5k870151
asked Nov 25 at 14:58
Rose
65
asked Nov 25 at 14:58
Rose
65
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:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
answered Nov 25 at 15:09
Hello_World
3,83721630
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 |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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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:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
answered Nov 25 at 15:09
Hello_World
3,83721630
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:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
answered Nov 25 at 15:09
Hello_World
3,83721630
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:
$$sum_{k=1}^{n}sin kx=frac{cosleft(frac{1}2xright)-cosleft(n+frac{1}2right)x}{2sin(x/2)},xneq0,pmpi,pm 2pi,...$$
answered Nov 25 at 15:09
Hello_World
3,83721630
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
3,83721630
answered Nov 25 at 15:09
Hello_World
3,83721630
answered Nov 25 at 15:09
Hello_World
3,83721630
answered Nov 25 at 15:09
Hello_World
3,83721630
3,83721630
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 |
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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