Mixing
Why this :$sumlimits_{n=1}^inftyfrac1{nsin n}$ is not harmonic series and is convergent?
2
$begingroup$
I'm confused how this series :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is a convergent series as wolfram show here and it's not harmonic series in the same time .
My question here is:
Why this :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is not harmonic series and is convergent ?
sequences-and-series convergence divergent-series
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
$endgroup$
|
show 14 more comments
2
$begingroup$
I'm confused how this series :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is a convergent series as wolfram show here and it's not harmonic series in the same time .
My question here is:
Why this :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is not harmonic series and is convergent ?
sequences-and-series convergence divergent-series
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
$endgroup$
2
$begingroup$
The harmonic series is $$sum_{n=1}^{infty} frac{1}{n}$$ and it diverges!
$endgroup$
– Peter
Nov 28 '16 at 21:33
4
$begingroup$
The series $$sum_{n = 1}^infty frac{1}{nsin n}$$ doesn't converge, its terms don't converge to $0$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:37
2
$begingroup$
If $frac{p}{q}$ is a convergent of $pi$, then $$biggllvert pi - frac{p}{q}biggrrvert < frac{1}{q^2}.$$ Then $lvert qpi - prvert < frac{1}{q}$ and hence $lvert sin prvert < frac{1}{q} approx frac{pi}{p}$. So for infinitely many $p$ (the numerators of the convergents of $pi$ we have $lvert psin prvert < 4$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:42
2
$begingroup$
@user51189 That still is wrong: the limit of $;frac1nsin n;$ is zero and still sine is periodic. The reasons you give are irrelevant for the existence of the limit and its value.
$endgroup$
– DonAntonio
Nov 28 '16 at 21:52
2
$begingroup$
@Peter We can easily show that $liminflimits_{ntoinfty} lvert nsin nrvert leqslant frac{pi}{sqrt{5}}$. On the other hand, clearly $limsuplimits_{ntoinfty} lvert nsin nrvert = +infty$. If the irrationality measure of $pi$ is strictly greater than $2$, we have $liminflimits_{ntoinfty} lvert nsin nrvert = 0$. We even have that if the irrationality measure of $pi$ is exactly $2$ and the sequence of partial quotients is unbounded. It is very likely that it is unbounded.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 22:34
|
show 14 more comments
2
2
2
1
$begingroup$
I'm confused how this series :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is a convergent series as wolfram show here and it's not harmonic series in the same time .
My question here is:
Why this :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is not harmonic series and is convergent ?
sequences-and-series convergence divergent-series
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
$endgroup$
I'm confused how this series :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is a convergent series as wolfram show here and it's not harmonic series in the same time .
My question here is:
Why this :$$displaystyle sum_{n=1}^inftyfrac1{nsin n}$$ is not harmonic series and is convergent ?
sequences-and-series convergence divergent-series
sequences-and-series convergence divergent-series
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
edited Jan 23 at 7:02
Martin Sleziak
44.8k10119273
44.8k10119273
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
asked Nov 28 '16 at 21:31
zeraoulia rafikzeraoulia rafik
2,40911031
2,40911031
2
$begingroup$
The harmonic series is $$sum_{n=1}^{infty} frac{1}{n}$$ and it diverges!
$endgroup$
– Peter
Nov 28 '16 at 21:33
4
$begingroup$
The series $$sum_{n = 1}^infty frac{1}{nsin n}$$ doesn't converge, its terms don't converge to $0$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:37
2
$begingroup$
If $frac{p}{q}$ is a convergent of $pi$, then $$biggllvert pi - frac{p}{q}biggrrvert < frac{1}{q^2}.$$ Then $lvert qpi - prvert < frac{1}{q}$ and hence $lvert sin prvert < frac{1}{q} approx frac{pi}{p}$. So for infinitely many $p$ (the numerators of the convergents of $pi$ we have $lvert psin prvert < 4$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:42
2
$begingroup$
@user51189 That still is wrong: the limit of $;frac1nsin n;$ is zero and still sine is periodic. The reasons you give are irrelevant for the existence of the limit and its value.
$endgroup$
– DonAntonio
Nov 28 '16 at 21:52
2
$begingroup$
@Peter We can easily show that $liminflimits_{ntoinfty} lvert nsin nrvert leqslant frac{pi}{sqrt{5}}$. On the other hand, clearly $limsuplimits_{ntoinfty} lvert nsin nrvert = +infty$. If the irrationality measure of $pi$ is strictly greater than $2$, we have $liminflimits_{ntoinfty} lvert nsin nrvert = 0$. We even have that if the irrationality measure of $pi$ is exactly $2$ and the sequence of partial quotients is unbounded. It is very likely that it is unbounded.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 22:34
|
show 14 more comments
2
$begingroup$
The harmonic series is $$sum_{n=1}^{infty} frac{1}{n}$$ and it diverges!
$endgroup$
– Peter
Nov 28 '16 at 21:33
4
$begingroup$
The series $$sum_{n = 1}^infty frac{1}{nsin n}$$ doesn't converge, its terms don't converge to $0$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:37
2
$begingroup$
If $frac{p}{q}$ is a convergent of $pi$, then $$biggllvert pi - frac{p}{q}biggrrvert < frac{1}{q^2}.$$ Then $lvert qpi - prvert < frac{1}{q}$ and hence $lvert sin prvert < frac{1}{q} approx frac{pi}{p}$. So for infinitely many $p$ (the numerators of the convergents of $pi$ we have $lvert psin prvert < 4$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:42
2
$begingroup$
@user51189 That still is wrong: the limit of $;frac1nsin n;$ is zero and still sine is periodic. The reasons you give are irrelevant for the existence of the limit and its value.
$endgroup$
– DonAntonio
Nov 28 '16 at 21:52
2
$begingroup$
@Peter We can easily show that $liminflimits_{ntoinfty} lvert nsin nrvert leqslant frac{pi}{sqrt{5}}$. On the other hand, clearly $limsuplimits_{ntoinfty} lvert nsin nrvert = +infty$. If the irrationality measure of $pi$ is strictly greater than $2$, we have $liminflimits_{ntoinfty} lvert nsin nrvert = 0$. We even have that if the irrationality measure of $pi$ is exactly $2$ and the sequence of partial quotients is unbounded. It is very likely that it is unbounded.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 22:34
2
2
$begingroup$
The harmonic series is $$sum_{n=1}^{infty} frac{1}{n}$$ and it diverges!
$endgroup$
– Peter
Nov 28 '16 at 21:33
$begingroup$
The harmonic series is $$sum_{n=1}^{infty} frac{1}{n}$$ and it diverges!
$endgroup$
– Peter
Nov 28 '16 at 21:33
4
4
$begingroup$
The series $$sum_{n = 1}^infty frac{1}{nsin n}$$ doesn't converge, its terms don't converge to $0$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:37
$begingroup$
The series $$sum_{n = 1}^infty frac{1}{nsin n}$$ doesn't converge, its terms don't converge to $0$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:37
2
2
$begingroup$
If $frac{p}{q}$ is a convergent of $pi$, then $$biggllvert pi - frac{p}{q}biggrrvert < frac{1}{q^2}.$$ Then $lvert qpi - prvert < frac{1}{q}$ and hence $lvert sin prvert < frac{1}{q} approx frac{pi}{p}$. So for infinitely many $p$ (the numerators of the convergents of $pi$ we have $lvert psin prvert < 4$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:42
$begingroup$
If $frac{p}{q}$ is a convergent of $pi$, then $$biggllvert pi - frac{p}{q}biggrrvert < frac{1}{q^2}.$$ Then $lvert qpi - prvert < frac{1}{q}$ and hence $lvert sin prvert < frac{1}{q} approx frac{pi}{p}$. So for infinitely many $p$ (the numerators of the convergents of $pi$ we have $lvert psin prvert < 4$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:42
2
2
$begingroup$
@user51189 That still is wrong: the limit of $;frac1nsin n;$ is zero and still sine is periodic. The reasons you give are irrelevant for the existence of the limit and its value.
$endgroup$
– DonAntonio
Nov 28 '16 at 21:52
$begingroup$
@user51189 That still is wrong: the limit of $;frac1nsin n;$ is zero and still sine is periodic. The reasons you give are irrelevant for the existence of the limit and its value.
$endgroup$
– DonAntonio
Nov 28 '16 at 21:52
2
2
$begingroup$
@Peter We can easily show that $liminflimits_{ntoinfty} lvert nsin nrvert leqslant frac{pi}{sqrt{5}}$. On the other hand, clearly $limsuplimits_{ntoinfty} lvert nsin nrvert = +infty$. If the irrationality measure of $pi$ is strictly greater than $2$, we have $liminflimits_{ntoinfty} lvert nsin nrvert = 0$. We even have that if the irrationality measure of $pi$ is exactly $2$ and the sequence of partial quotients is unbounded. It is very likely that it is unbounded.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 22:34
$begingroup$
@Peter We can easily show that $liminflimits_{ntoinfty} lvert nsin nrvert leqslant frac{pi}{sqrt{5}}$. On the other hand, clearly $limsuplimits_{ntoinfty} lvert nsin nrvert = +infty$. If the irrationality measure of $pi$ is strictly greater than $2$, we have $liminflimits_{ntoinfty} lvert nsin nrvert = 0$. We even have that if the irrationality measure of $pi$ is exactly $2$ and the sequence of partial quotients is unbounded. It is very likely that it is unbounded.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 22:34
|
show 14 more comments
1 Answer
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Hint: According to the answer by @J.R. given at this question the sequence
begin{align*}
left(frac{1}{nsin n}right)_{ngeq 1}
end{align*}
is not convergent. We conclude the series
begin{align*}
sum_{n=1}^inftyfrac{1}{nsin n}
end{align*}
is divergent.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
add a comment |
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$begingroup$
Hint: According to the answer by @J.R. given at this question the sequence
begin{align*}
left(frac{1}{nsin n}right)_{ngeq 1}
end{align*}
is not convergent. We conclude the series
begin{align*}
sum_{n=1}^inftyfrac{1}{nsin n}
end{align*}
is divergent.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
add a comment |
1
$begingroup$
Hint: According to the answer by @J.R. given at this question the sequence
begin{align*}
left(frac{1}{nsin n}right)_{ngeq 1}
end{align*}
is not convergent. We conclude the series
begin{align*}
sum_{n=1}^inftyfrac{1}{nsin n}
end{align*}
is divergent.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
add a comment |
1
1
1
$begingroup$
Hint: According to the answer by @J.R. given at this question the sequence
begin{align*}
left(frac{1}{nsin n}right)_{ngeq 1}
end{align*}
is not convergent. We conclude the series
begin{align*}
sum_{n=1}^inftyfrac{1}{nsin n}
end{align*}
is divergent.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
$endgroup$
Hint: According to the answer by @J.R. given at this question the sequence
begin{align*}
left(frac{1}{nsin n}right)_{ngeq 1}
end{align*}
is not convergent. We conclude the series
begin{align*}
sum_{n=1}^inftyfrac{1}{nsin n}
end{align*}
is divergent.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
answered Nov 29 '16 at 16:49
Markus ScheuerMarkus Scheuer
62.6k459149
62.6k459149
add a comment |
add a comment |
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2
$begingroup$
The harmonic series is $$sum_{n=1}^{infty} frac{1}{n}$$ and it diverges!
$endgroup$
– Peter
Nov 28 '16 at 21:33
4
$begingroup$
The series $$sum_{n = 1}^infty frac{1}{nsin n}$$ doesn't converge, its terms don't converge to $0$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:37
2
$begingroup$
If $frac{p}{q}$ is a convergent of $pi$, then $$biggllvert pi - frac{p}{q}biggrrvert < frac{1}{q^2}.$$ Then $lvert qpi - prvert < frac{1}{q}$ and hence $lvert sin prvert < frac{1}{q} approx frac{pi}{p}$. So for infinitely many $p$ (the numerators of the convergents of $pi$ we have $lvert psin prvert < 4$.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 21:42
2
$begingroup$
@user51189 That still is wrong: the limit of $;frac1nsin n;$ is zero and still sine is periodic. The reasons you give are irrelevant for the existence of the limit and its value.
$endgroup$
– DonAntonio
Nov 28 '16 at 21:52
2
$begingroup$
@Peter We can easily show that $liminflimits_{ntoinfty} lvert nsin nrvert leqslant frac{pi}{sqrt{5}}$. On the other hand, clearly $limsuplimits_{ntoinfty} lvert nsin nrvert = +infty$. If the irrationality measure of $pi$ is strictly greater than $2$, we have $liminflimits_{ntoinfty} lvert nsin nrvert = 0$. We even have that if the irrationality measure of $pi$ is exactly $2$ and the sequence of partial quotients is unbounded. It is very likely that it is unbounded.
$endgroup$
– Daniel Fischer♦
Nov 28 '16 at 22:34