Convergence of $lim_{Ntoinfty}sum_{n=1}^N exp(-Nsin^2(frac{npi}{2N}))$ and $lim_{Ntoinfty}sum_{n=1}^N...











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down vote

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I can't find the right approach to tackle the question whether
$$lim_{Ntoinfty} sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)$$
and
$$lim_{Ntoinfty} sum_{n=1}^N expBiggl(-sin^2left(frac{npi}{2N}right)Biggr)$$
converge or diverge. The fact that the limiting variable appears both as the upper bound of summation as well as in the individual summands seems to make the standard methods known to me inapplicable.



I suspect that the second limit (i.e. the one not containing $N$ in the exponent directly) does not exist, but that the first one may. I would be very grateful if you could point me to methods that allow one to determine the existence of the limits.



If a limit exists, I would also be very interested in understanding how, if at all, one could (approximately) replace the sum with an integral.



(For background, these questions have arisen during my study of the Rouse theory of polymer dynamics, e.g. in chapter 7.3.2. of Doi and Edwards, "The Theory of Polymer Dynamics". Physical explanations of how one can justify the treatment therein would be very welcome, too.)



Thank you in advance!










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  • ru.wikipedia.org/wiki/Логарифмический_признак_сходимости
    – Samvel Safaryan
    Nov 21 at 13:35






  • 1




    @SamvelSafaryan: Thanks for the link. I'm afraid, though, that this method is not applicable, or is it? I assume the theorem holds only if the $a_n$ do not depend on the limiting variable $N$ (i.e. the maximum index of summation). Please correct me if I'm wrong, I might just have conceptual difficulties.
    – Batista
    Nov 21 at 13:55










  • The second sum diverges to $infty $ because the same sum divided by $N$ tends to $int_{0}^{1}e^{-sin^2(pi x/2)},dx>0$.
    – Paramanand Singh
    Nov 21 at 14:21












  • @ParamanandSingh: That's an elegant argument, thanks!
    – Batista
    Nov 21 at 14:47






  • 2




    Replacing $e^{-Ncdots}$ with $e^{-N^2cdots}$ gives a more interesting problem with the limit converging to a finite value.
    – Winther
    Nov 21 at 23:40

















up vote
3
down vote

favorite
1












I can't find the right approach to tackle the question whether
$$lim_{Ntoinfty} sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)$$
and
$$lim_{Ntoinfty} sum_{n=1}^N expBiggl(-sin^2left(frac{npi}{2N}right)Biggr)$$
converge or diverge. The fact that the limiting variable appears both as the upper bound of summation as well as in the individual summands seems to make the standard methods known to me inapplicable.



I suspect that the second limit (i.e. the one not containing $N$ in the exponent directly) does not exist, but that the first one may. I would be very grateful if you could point me to methods that allow one to determine the existence of the limits.



If a limit exists, I would also be very interested in understanding how, if at all, one could (approximately) replace the sum with an integral.



(For background, these questions have arisen during my study of the Rouse theory of polymer dynamics, e.g. in chapter 7.3.2. of Doi and Edwards, "The Theory of Polymer Dynamics". Physical explanations of how one can justify the treatment therein would be very welcome, too.)



Thank you in advance!










share|cite|improve this question









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Batista is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • ru.wikipedia.org/wiki/Логарифмический_признак_сходимости
    – Samvel Safaryan
    Nov 21 at 13:35






  • 1




    @SamvelSafaryan: Thanks for the link. I'm afraid, though, that this method is not applicable, or is it? I assume the theorem holds only if the $a_n$ do not depend on the limiting variable $N$ (i.e. the maximum index of summation). Please correct me if I'm wrong, I might just have conceptual difficulties.
    – Batista
    Nov 21 at 13:55










  • The second sum diverges to $infty $ because the same sum divided by $N$ tends to $int_{0}^{1}e^{-sin^2(pi x/2)},dx>0$.
    – Paramanand Singh
    Nov 21 at 14:21












  • @ParamanandSingh: That's an elegant argument, thanks!
    – Batista
    Nov 21 at 14:47






  • 2




    Replacing $e^{-Ncdots}$ with $e^{-N^2cdots}$ gives a more interesting problem with the limit converging to a finite value.
    – Winther
    Nov 21 at 23:40















up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I can't find the right approach to tackle the question whether
$$lim_{Ntoinfty} sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)$$
and
$$lim_{Ntoinfty} sum_{n=1}^N expBiggl(-sin^2left(frac{npi}{2N}right)Biggr)$$
converge or diverge. The fact that the limiting variable appears both as the upper bound of summation as well as in the individual summands seems to make the standard methods known to me inapplicable.



I suspect that the second limit (i.e. the one not containing $N$ in the exponent directly) does not exist, but that the first one may. I would be very grateful if you could point me to methods that allow one to determine the existence of the limits.



If a limit exists, I would also be very interested in understanding how, if at all, one could (approximately) replace the sum with an integral.



(For background, these questions have arisen during my study of the Rouse theory of polymer dynamics, e.g. in chapter 7.3.2. of Doi and Edwards, "The Theory of Polymer Dynamics". Physical explanations of how one can justify the treatment therein would be very welcome, too.)



Thank you in advance!










share|cite|improve this question









New contributor




Batista is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I can't find the right approach to tackle the question whether
$$lim_{Ntoinfty} sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)$$
and
$$lim_{Ntoinfty} sum_{n=1}^N expBiggl(-sin^2left(frac{npi}{2N}right)Biggr)$$
converge or diverge. The fact that the limiting variable appears both as the upper bound of summation as well as in the individual summands seems to make the standard methods known to me inapplicable.



I suspect that the second limit (i.e. the one not containing $N$ in the exponent directly) does not exist, but that the first one may. I would be very grateful if you could point me to methods that allow one to determine the existence of the limits.



If a limit exists, I would also be very interested in understanding how, if at all, one could (approximately) replace the sum with an integral.



(For background, these questions have arisen during my study of the Rouse theory of polymer dynamics, e.g. in chapter 7.3.2. of Doi and Edwards, "The Theory of Polymer Dynamics". Physical explanations of how one can justify the treatment therein would be very welcome, too.)



Thank you in advance!







sequences-and-series limits convergence summation exponential-sum






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edited Nov 21 at 16:58





















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asked Nov 21 at 12:29









Batista

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  • ru.wikipedia.org/wiki/Логарифмический_признак_сходимости
    – Samvel Safaryan
    Nov 21 at 13:35






  • 1




    @SamvelSafaryan: Thanks for the link. I'm afraid, though, that this method is not applicable, or is it? I assume the theorem holds only if the $a_n$ do not depend on the limiting variable $N$ (i.e. the maximum index of summation). Please correct me if I'm wrong, I might just have conceptual difficulties.
    – Batista
    Nov 21 at 13:55










  • The second sum diverges to $infty $ because the same sum divided by $N$ tends to $int_{0}^{1}e^{-sin^2(pi x/2)},dx>0$.
    – Paramanand Singh
    Nov 21 at 14:21












  • @ParamanandSingh: That's an elegant argument, thanks!
    – Batista
    Nov 21 at 14:47






  • 2




    Replacing $e^{-Ncdots}$ with $e^{-N^2cdots}$ gives a more interesting problem with the limit converging to a finite value.
    – Winther
    Nov 21 at 23:40




















  • ru.wikipedia.org/wiki/Логарифмический_признак_сходимости
    – Samvel Safaryan
    Nov 21 at 13:35






  • 1




    @SamvelSafaryan: Thanks for the link. I'm afraid, though, that this method is not applicable, or is it? I assume the theorem holds only if the $a_n$ do not depend on the limiting variable $N$ (i.e. the maximum index of summation). Please correct me if I'm wrong, I might just have conceptual difficulties.
    – Batista
    Nov 21 at 13:55










  • The second sum diverges to $infty $ because the same sum divided by $N$ tends to $int_{0}^{1}e^{-sin^2(pi x/2)},dx>0$.
    – Paramanand Singh
    Nov 21 at 14:21












  • @ParamanandSingh: That's an elegant argument, thanks!
    – Batista
    Nov 21 at 14:47






  • 2




    Replacing $e^{-Ncdots}$ with $e^{-N^2cdots}$ gives a more interesting problem with the limit converging to a finite value.
    – Winther
    Nov 21 at 23:40


















ru.wikipedia.org/wiki/Логарифмический_признак_сходимости
– Samvel Safaryan
Nov 21 at 13:35




ru.wikipedia.org/wiki/Логарифмический_признак_сходимости
– Samvel Safaryan
Nov 21 at 13:35




1




1




@SamvelSafaryan: Thanks for the link. I'm afraid, though, that this method is not applicable, or is it? I assume the theorem holds only if the $a_n$ do not depend on the limiting variable $N$ (i.e. the maximum index of summation). Please correct me if I'm wrong, I might just have conceptual difficulties.
– Batista
Nov 21 at 13:55




@SamvelSafaryan: Thanks for the link. I'm afraid, though, that this method is not applicable, or is it? I assume the theorem holds only if the $a_n$ do not depend on the limiting variable $N$ (i.e. the maximum index of summation). Please correct me if I'm wrong, I might just have conceptual difficulties.
– Batista
Nov 21 at 13:55












The second sum diverges to $infty $ because the same sum divided by $N$ tends to $int_{0}^{1}e^{-sin^2(pi x/2)},dx>0$.
– Paramanand Singh
Nov 21 at 14:21






The second sum diverges to $infty $ because the same sum divided by $N$ tends to $int_{0}^{1}e^{-sin^2(pi x/2)},dx>0$.
– Paramanand Singh
Nov 21 at 14:21














@ParamanandSingh: That's an elegant argument, thanks!
– Batista
Nov 21 at 14:47




@ParamanandSingh: That's an elegant argument, thanks!
– Batista
Nov 21 at 14:47




2




2




Replacing $e^{-Ncdots}$ with $e^{-N^2cdots}$ gives a more interesting problem with the limit converging to a finite value.
– Winther
Nov 21 at 23:40






Replacing $e^{-Ncdots}$ with $e^{-N^2cdots}$ gives a more interesting problem with the limit converging to a finite value.
– Winther
Nov 21 at 23:40












2 Answers
2






active

oldest

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up vote
3
down vote



accepted










Here's another way to see the limit of the first series is $infty.$ Verify that for each fixed $n,$



$$lim_{Nto infty}expleft(-N sin^2left(frac{npi}{2N}right)right) = 1.$$



Fix $N_0in mathbb N.$ Then for $Nge N_0,$ the first series is at least



$$sum_{n=1}^{N_0}expleft(-N sin^2left(frac{npi}{2N}right)right).$$



This is a finite sum, so computing its limit as $Nto infty$ is easy: we get $sum_{n=1}^{N_0}1= N_0.$ Since $N_0$ is arbitrarily large, the limit of the first series is $infty.$ Since the terms of the second series are at least as large as the those of the first series, the limit for the second series is also $infty.$






share|cite|improve this answer























  • Thank you very much, that's a very elegant solution!
    – Batista
    2 days ago


















up vote
3
down vote













Both sums diverge and a general idea to prove it is to use Riemann sums, just as suggested in the comments. The second sum was proved divergent in the comments. For the first sum notice that $0 le sin x le x$ for $x in [0, pi/2]$ implies $0 geq - sin^2 x geq -x^2$ in this range and since the exponential function is increasing we have
begin{align*}
sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)
&geq sum_{n=1}^N expBigg(- frac{n^2pi^2}{4N}Bigg) \
&= sqrt{N} cdot left[ frac{1}{sqrt{N} } sum_{n=1}^N expBigg(- frac{pi^2}{4} bigg(frac{n}{sqrt{N}}bigg)^2 Bigg)right]
end{align*}

Now, the expression inside square brackets is close to the integral (you can compare them using integral test for instance):
$$
int_0^{sqrt{N}} e^{-frac{pi^2}{4} x^2} dx to int_0^infty e^{-frac{pi^2}{4} x^2} dx < infty quad text{as $N to infty$}.
$$

This implies the first integral grows at least as a multiple of $sqrt{N}$ and must diverge.






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  • Thank you so much for your insightful answer! I wish I could accept both answers.
    – Batista
    2 days ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Here's another way to see the limit of the first series is $infty.$ Verify that for each fixed $n,$



$$lim_{Nto infty}expleft(-N sin^2left(frac{npi}{2N}right)right) = 1.$$



Fix $N_0in mathbb N.$ Then for $Nge N_0,$ the first series is at least



$$sum_{n=1}^{N_0}expleft(-N sin^2left(frac{npi}{2N}right)right).$$



This is a finite sum, so computing its limit as $Nto infty$ is easy: we get $sum_{n=1}^{N_0}1= N_0.$ Since $N_0$ is arbitrarily large, the limit of the first series is $infty.$ Since the terms of the second series are at least as large as the those of the first series, the limit for the second series is also $infty.$






share|cite|improve this answer























  • Thank you very much, that's a very elegant solution!
    – Batista
    2 days ago















up vote
3
down vote



accepted










Here's another way to see the limit of the first series is $infty.$ Verify that for each fixed $n,$



$$lim_{Nto infty}expleft(-N sin^2left(frac{npi}{2N}right)right) = 1.$$



Fix $N_0in mathbb N.$ Then for $Nge N_0,$ the first series is at least



$$sum_{n=1}^{N_0}expleft(-N sin^2left(frac{npi}{2N}right)right).$$



This is a finite sum, so computing its limit as $Nto infty$ is easy: we get $sum_{n=1}^{N_0}1= N_0.$ Since $N_0$ is arbitrarily large, the limit of the first series is $infty.$ Since the terms of the second series are at least as large as the those of the first series, the limit for the second series is also $infty.$






share|cite|improve this answer























  • Thank you very much, that's a very elegant solution!
    – Batista
    2 days ago













up vote
3
down vote



accepted







up vote
3
down vote



accepted






Here's another way to see the limit of the first series is $infty.$ Verify that for each fixed $n,$



$$lim_{Nto infty}expleft(-N sin^2left(frac{npi}{2N}right)right) = 1.$$



Fix $N_0in mathbb N.$ Then for $Nge N_0,$ the first series is at least



$$sum_{n=1}^{N_0}expleft(-N sin^2left(frac{npi}{2N}right)right).$$



This is a finite sum, so computing its limit as $Nto infty$ is easy: we get $sum_{n=1}^{N_0}1= N_0.$ Since $N_0$ is arbitrarily large, the limit of the first series is $infty.$ Since the terms of the second series are at least as large as the those of the first series, the limit for the second series is also $infty.$






share|cite|improve this answer














Here's another way to see the limit of the first series is $infty.$ Verify that for each fixed $n,$



$$lim_{Nto infty}expleft(-N sin^2left(frac{npi}{2N}right)right) = 1.$$



Fix $N_0in mathbb N.$ Then for $Nge N_0,$ the first series is at least



$$sum_{n=1}^{N_0}expleft(-N sin^2left(frac{npi}{2N}right)right).$$



This is a finite sum, so computing its limit as $Nto infty$ is easy: we get $sum_{n=1}^{N_0}1= N_0.$ Since $N_0$ is arbitrarily large, the limit of the first series is $infty.$ Since the terms of the second series are at least as large as the those of the first series, the limit for the second series is also $infty.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 23:28

























answered Nov 21 at 21:02









zhw.

70.4k42975




70.4k42975












  • Thank you very much, that's a very elegant solution!
    – Batista
    2 days ago


















  • Thank you very much, that's a very elegant solution!
    – Batista
    2 days ago
















Thank you very much, that's a very elegant solution!
– Batista
2 days ago




Thank you very much, that's a very elegant solution!
– Batista
2 days ago










up vote
3
down vote













Both sums diverge and a general idea to prove it is to use Riemann sums, just as suggested in the comments. The second sum was proved divergent in the comments. For the first sum notice that $0 le sin x le x$ for $x in [0, pi/2]$ implies $0 geq - sin^2 x geq -x^2$ in this range and since the exponential function is increasing we have
begin{align*}
sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)
&geq sum_{n=1}^N expBigg(- frac{n^2pi^2}{4N}Bigg) \
&= sqrt{N} cdot left[ frac{1}{sqrt{N} } sum_{n=1}^N expBigg(- frac{pi^2}{4} bigg(frac{n}{sqrt{N}}bigg)^2 Bigg)right]
end{align*}

Now, the expression inside square brackets is close to the integral (you can compare them using integral test for instance):
$$
int_0^{sqrt{N}} e^{-frac{pi^2}{4} x^2} dx to int_0^infty e^{-frac{pi^2}{4} x^2} dx < infty quad text{as $N to infty$}.
$$

This implies the first integral grows at least as a multiple of $sqrt{N}$ and must diverge.






share|cite|improve this answer





















  • Thank you so much for your insightful answer! I wish I could accept both answers.
    – Batista
    2 days ago















up vote
3
down vote













Both sums diverge and a general idea to prove it is to use Riemann sums, just as suggested in the comments. The second sum was proved divergent in the comments. For the first sum notice that $0 le sin x le x$ for $x in [0, pi/2]$ implies $0 geq - sin^2 x geq -x^2$ in this range and since the exponential function is increasing we have
begin{align*}
sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)
&geq sum_{n=1}^N expBigg(- frac{n^2pi^2}{4N}Bigg) \
&= sqrt{N} cdot left[ frac{1}{sqrt{N} } sum_{n=1}^N expBigg(- frac{pi^2}{4} bigg(frac{n}{sqrt{N}}bigg)^2 Bigg)right]
end{align*}

Now, the expression inside square brackets is close to the integral (you can compare them using integral test for instance):
$$
int_0^{sqrt{N}} e^{-frac{pi^2}{4} x^2} dx to int_0^infty e^{-frac{pi^2}{4} x^2} dx < infty quad text{as $N to infty$}.
$$

This implies the first integral grows at least as a multiple of $sqrt{N}$ and must diverge.






share|cite|improve this answer





















  • Thank you so much for your insightful answer! I wish I could accept both answers.
    – Batista
    2 days ago













up vote
3
down vote










up vote
3
down vote









Both sums diverge and a general idea to prove it is to use Riemann sums, just as suggested in the comments. The second sum was proved divergent in the comments. For the first sum notice that $0 le sin x le x$ for $x in [0, pi/2]$ implies $0 geq - sin^2 x geq -x^2$ in this range and since the exponential function is increasing we have
begin{align*}
sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)
&geq sum_{n=1}^N expBigg(- frac{n^2pi^2}{4N}Bigg) \
&= sqrt{N} cdot left[ frac{1}{sqrt{N} } sum_{n=1}^N expBigg(- frac{pi^2}{4} bigg(frac{n}{sqrt{N}}bigg)^2 Bigg)right]
end{align*}

Now, the expression inside square brackets is close to the integral (you can compare them using integral test for instance):
$$
int_0^{sqrt{N}} e^{-frac{pi^2}{4} x^2} dx to int_0^infty e^{-frac{pi^2}{4} x^2} dx < infty quad text{as $N to infty$}.
$$

This implies the first integral grows at least as a multiple of $sqrt{N}$ and must diverge.






share|cite|improve this answer












Both sums diverge and a general idea to prove it is to use Riemann sums, just as suggested in the comments. The second sum was proved divergent in the comments. For the first sum notice that $0 le sin x le x$ for $x in [0, pi/2]$ implies $0 geq - sin^2 x geq -x^2$ in this range and since the exponential function is increasing we have
begin{align*}
sum_{n=1}^N expBigg(-N sin^2left(frac{npi}{2N}right)Bigg)
&geq sum_{n=1}^N expBigg(- frac{n^2pi^2}{4N}Bigg) \
&= sqrt{N} cdot left[ frac{1}{sqrt{N} } sum_{n=1}^N expBigg(- frac{pi^2}{4} bigg(frac{n}{sqrt{N}}bigg)^2 Bigg)right]
end{align*}

Now, the expression inside square brackets is close to the integral (you can compare them using integral test for instance):
$$
int_0^{sqrt{N}} e^{-frac{pi^2}{4} x^2} dx to int_0^infty e^{-frac{pi^2}{4} x^2} dx < infty quad text{as $N to infty$}.
$$

This implies the first integral grows at least as a multiple of $sqrt{N}$ and must diverge.







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answered Nov 21 at 17:48









Daniel

1,22628




1,22628












  • Thank you so much for your insightful answer! I wish I could accept both answers.
    – Batista
    2 days ago


















  • Thank you so much for your insightful answer! I wish I could accept both answers.
    – Batista
    2 days ago
















Thank you so much for your insightful answer! I wish I could accept both answers.
– Batista
2 days ago




Thank you so much for your insightful answer! I wish I could accept both answers.
– Batista
2 days ago










Batista is a new contributor. Be nice, and check out our Code of Conduct.










 

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