Does $sumlimits_n=1^inftysin(n)sinleft(fracpi2nright)$ converge?
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I must determine whether if the following series converges, converges absolutely, or diverges: $$sum_n=1^inftysin(n)sinleft(fracpi2nright)$$ By the comparison test, I have already found that $sumlimits_n=1^infty left(sinleft(fracpi2nright)right)^p$ converges for $p>1$ and diverges for $p leq 1$. Thus, $
sumlimits_n=1^inftysinleft(fracpi2nright)$ diverges by this criterion. I suspect the entire series will also diverge, and that I have to use the comparison test, but I encountered an issue:
Since $-1 leq sin n leq 1$, we have that $sin(n)sinleft(fracpi2nright) leq sinleft(fracpi2nright)$. This would be useful if the series represented by the term on the right converged; in its current state, this cannot be used to prove divergence.
Is my reasoning wrong? Should I be using another test for this series? Thank you.
calculus real-analysis sequences-and-series limits convergence
edited Aug 30 at 5:50
Robson
47320
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
add a comment |Â
up vote
8
down vote
favorite
I must determine whether if the following series converges, converges absolutely, or diverges: $$sum_n=1^inftysin(n)sinleft(fracpi2nright)$$ By the comparison test, I have already found that $sumlimits_n=1^infty left(sinleft(fracpi2nright)right)^p$ converges for $p>1$ and diverges for $p leq 1$. Thus, $
sumlimits_n=1^inftysinleft(fracpi2nright)$ diverges by this criterion. I suspect the entire series will also diverge, and that I have to use the comparison test, but I encountered an issue:
Since $-1 leq sin n leq 1$, we have that $sin(n)sinleft(fracpi2nright) leq sinleft(fracpi2nright)$. This would be useful if the series represented by the term on the right converged; in its current state, this cannot be used to prove divergence.
Is my reasoning wrong? Should I be using another test for this series? Thank you.
calculus real-analysis sequences-and-series limits convergence
edited Aug 30 at 5:50
Robson
47320
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
1
It certainly can't converge absolutely (and that should be straightforward to prove), but convergence of this series looks remarkably tricky at least at first glance...
– Steven Stadnicki
Apr 1 '15 at 3:48
3
Dirichlet test does the trick.
– Gabriel Romon
Apr 1 '15 at 3:49
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
I must determine whether if the following series converges, converges absolutely, or diverges: $$sum_n=1^inftysin(n)sinleft(fracpi2nright)$$ By the comparison test, I have already found that $sumlimits_n=1^infty left(sinleft(fracpi2nright)right)^p$ converges for $p>1$ and diverges for $p leq 1$. Thus, $
sumlimits_n=1^inftysinleft(fracpi2nright)$ diverges by this criterion. I suspect the entire series will also diverge, and that I have to use the comparison test, but I encountered an issue:
Since $-1 leq sin n leq 1$, we have that $sin(n)sinleft(fracpi2nright) leq sinleft(fracpi2nright)$. This would be useful if the series represented by the term on the right converged; in its current state, this cannot be used to prove divergence.
Is my reasoning wrong? Should I be using another test for this series? Thank you.
calculus real-analysis sequences-and-series limits convergence
edited Aug 30 at 5:50
Robson
47320
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
I must determine whether if the following series converges, converges absolutely, or diverges: $$sum_n=1^inftysin(n)sinleft(fracpi2nright)$$ By the comparison test, I have already found that $sumlimits_n=1^infty left(sinleft(fracpi2nright)right)^p$ converges for $p>1$ and diverges for $p leq 1$. Thus, $
sumlimits_n=1^inftysinleft(fracpi2nright)$ diverges by this criterion. I suspect the entire series will also diverge, and that I have to use the comparison test, but I encountered an issue:
Since $-1 leq sin n leq 1$, we have that $sin(n)sinleft(fracpi2nright) leq sinleft(fracpi2nright)$. This would be useful if the series represented by the term on the right converged; in its current state, this cannot be used to prove divergence.
Is my reasoning wrong? Should I be using another test for this series? Thank you.
calculus real-analysis sequences-and-series limits convergence
calculus real-analysis sequences-and-series limits convergence
edited Aug 30 at 5:50
Robson
47320
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
edited Aug 30 at 5:50
Robson
47320
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
edited Aug 30 at 5:50
Robson
47320
edited Aug 30 at 5:50
Robson
47320
edited Aug 30 at 5:50
Robson
47320
47320
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
asked Apr 1 '15 at 3:43
Douglas Fir
1,196818
1,196818
1
It certainly can't converge absolutely (and that should be straightforward to prove), but convergence of this series looks remarkably tricky at least at first glance...
– Steven Stadnicki
Apr 1 '15 at 3:48
3
Dirichlet test does the trick.
– Gabriel Romon
Apr 1 '15 at 3:49
add a comment |Â
1
It certainly can't converge absolutely (and that should be straightforward to prove), but convergence of this series looks remarkably tricky at least at first glance...
– Steven Stadnicki
Apr 1 '15 at 3:48
3
Dirichlet test does the trick.
– Gabriel Romon
Apr 1 '15 at 3:49
1
1
It certainly can't converge absolutely (and that should be straightforward to prove), but convergence of this series looks remarkably tricky at least at first glance...
– Steven Stadnicki
Apr 1 '15 at 3:48
It certainly can't converge absolutely (and that should be straightforward to prove), but convergence of this series looks remarkably tricky at least at first glance...
– Steven Stadnicki
Apr 1 '15 at 3:48
3
3
Dirichlet test does the trick.
– Gabriel Romon
Apr 1 '15 at 3:49
Dirichlet test does the trick.
– Gabriel Romon
Apr 1 '15 at 3:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Since the partial sums of $sum_nsin(n)$ are bounded and $sin(fracpi2n)$ is a decreasing sequence that goes to $0$, Dirichlet test proves the convergence of your series.
To get a bound on the partial sums of $sum_nsin(n)$, note that $displaystyle left|sum_k=0^n sin(k)right|=left|Imsum_k=0^ne^ikright|leqleft|frac1-e^i(n+1)1-e^iright|leq frac21-e^i$
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
How would I now prove that the series is not absolutely convergent? For $sum_n=1^infty lvert sin n rvert sinleft(fracpi2nright)$, we have that $lvert sin n rvert sinleft(fracpi2nright) leq sinleft(fracpi2nright)$, but again, this cannot be used.
– Douglas Fir
Apr 1 '15 at 4:43
1
This is a different question. Use the inequality $forall xin [-pi/2,pi/2], |sin(x)|geq frac2pi|x|$ and the divergence of the series $sum fracn$ (see math.stackexchange.com/questions/264980/…)
– Gabriel Romon
Apr 1 '15 at 4:47
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Since the partial sums of $sum_nsin(n)$ are bounded and $sin(fracpi2n)$ is a decreasing sequence that goes to $0$, Dirichlet test proves the convergence of your series.
To get a bound on the partial sums of $sum_nsin(n)$, note that $displaystyle left|sum_k=0^n sin(k)right|=left|Imsum_k=0^ne^ikright|leqleft|frac1-e^i(n+1)1-e^iright|leq frac21-e^i$
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
How would I now prove that the series is not absolutely convergent? For $sum_n=1^infty lvert sin n rvert sinleft(fracpi2nright)$, we have that $lvert sin n rvert sinleft(fracpi2nright) leq sinleft(fracpi2nright)$, but again, this cannot be used.
– Douglas Fir
Apr 1 '15 at 4:43
1
This is a different question. Use the inequality $forall xin [-pi/2,pi/2], |sin(x)|geq frac2pi|x|$ and the divergence of the series $sum fracn$ (see math.stackexchange.com/questions/264980/…)
– Gabriel Romon
Apr 1 '15 at 4:47
add a comment |Â
up vote
6
down vote
accepted
Since the partial sums of $sum_nsin(n)$ are bounded and $sin(fracpi2n)$ is a decreasing sequence that goes to $0$, Dirichlet test proves the convergence of your series.
To get a bound on the partial sums of $sum_nsin(n)$, note that $displaystyle left|sum_k=0^n sin(k)right|=left|Imsum_k=0^ne^ikright|leqleft|frac1-e^i(n+1)1-e^iright|leq frac21-e^i$
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
How would I now prove that the series is not absolutely convergent? For $sum_n=1^infty lvert sin n rvert sinleft(fracpi2nright)$, we have that $lvert sin n rvert sinleft(fracpi2nright) leq sinleft(fracpi2nright)$, but again, this cannot be used.
– Douglas Fir
Apr 1 '15 at 4:43
1
This is a different question. Use the inequality $forall xin [-pi/2,pi/2], |sin(x)|geq frac2pi|x|$ and the divergence of the series $sum fracn$ (see math.stackexchange.com/questions/264980/…)
– Gabriel Romon
Apr 1 '15 at 4:47
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Since the partial sums of $sum_nsin(n)$ are bounded and $sin(fracpi2n)$ is a decreasing sequence that goes to $0$, Dirichlet test proves the convergence of your series.
To get a bound on the partial sums of $sum_nsin(n)$, note that $displaystyle left|sum_k=0^n sin(k)right|=left|Imsum_k=0^ne^ikright|leqleft|frac1-e^i(n+1)1-e^iright|leq frac21-e^i$
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
Since the partial sums of $sum_nsin(n)$ are bounded and $sin(fracpi2n)$ is a decreasing sequence that goes to $0$, Dirichlet test proves the convergence of your series.
To get a bound on the partial sums of $sum_nsin(n)$, note that $displaystyle left|sum_k=0^n sin(k)right|=left|Imsum_k=0^ne^ikright|leqleft|frac1-e^i(n+1)1-e^iright|leq frac21-e^i$
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
edited Apr 1 '15 at 4:01
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
answered Apr 1 '15 at 3:53
Gabriel Romon
17.4k53182
17.4k53182
How would I now prove that the series is not absolutely convergent? For $sum_n=1^infty lvert sin n rvert sinleft(fracpi2nright)$, we have that $lvert sin n rvert sinleft(fracpi2nright) leq sinleft(fracpi2nright)$, but again, this cannot be used.
– Douglas Fir
Apr 1 '15 at 4:43
1
This is a different question. Use the inequality $forall xin [-pi/2,pi/2], |sin(x)|geq frac2pi|x|$ and the divergence of the series $sum fracn$ (see math.stackexchange.com/questions/264980/…)
– Gabriel Romon
Apr 1 '15 at 4:47
add a comment |Â
How would I now prove that the series is not absolutely convergent? For $sum_n=1^infty lvert sin n rvert sinleft(fracpi2nright)$, we have that $lvert sin n rvert sinleft(fracpi2nright) leq sinleft(fracpi2nright)$, but again, this cannot be used.
– Douglas Fir
Apr 1 '15 at 4:43
1
This is a different question. Use the inequality $forall xin [-pi/2,pi/2], |sin(x)|geq frac2pi|x|$ and the divergence of the series $sum fracn$ (see math.stackexchange.com/questions/264980/…)
– Gabriel Romon
Apr 1 '15 at 4:47
How would I now prove that the series is not absolutely convergent? For $sum_n=1^infty lvert sin n rvert sinleft(fracpi2nright)$, we have that $lvert sin n rvert sinleft(fracpi2nright) leq sinleft(fracpi2nright)$, but again, this cannot be used.
– Douglas Fir
Apr 1 '15 at 4:43
How would I now prove that the series is not absolutely convergent? For $sum_n=1^infty lvert sin n rvert sinleft(fracpi2nright)$, we have that $lvert sin n rvert sinleft(fracpi2nright) leq sinleft(fracpi2nright)$, but again, this cannot be used.
– Douglas Fir
Apr 1 '15 at 4:43
1
1
This is a different question. Use the inequality $forall xin [-pi/2,pi/2], |sin(x)|geq frac2pi|x|$ and the divergence of the series $sum fracn$ (see math.stackexchange.com/questions/264980/…)
– Gabriel Romon
Apr 1 '15 at 4:47
This is a different question. Use the inequality $forall xin [-pi/2,pi/2], |sin(x)|geq frac2pi|x|$ and the divergence of the series $sum fracn$ (see math.stackexchange.com/questions/264980/…)
– Gabriel Romon
Apr 1 '15 at 4:47
add a comment |Â
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1
It certainly can't converge absolutely (and that should be straightforward to prove), but convergence of this series looks remarkably tricky at least at first glance...
– Steven Stadnicki
Apr 1 '15 at 3:48
3
Dirichlet test does the trick.
– Gabriel Romon
Apr 1 '15 at 3:49