...












7












$begingroup$


I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.



Proof:



We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$






sequences-and-series proof-verification fourier-series decimal-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is a great way to obtain some nice problems for math contests or exams :D
    $endgroup$
    – Zacky
    Dec 25 '18 at 1:35






  • 1




    $begingroup$
    Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
    $endgroup$
    – Kemono Chen
    Dec 25 '18 at 1:50






  • 2




    $begingroup$
    @Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
    $endgroup$
    – clathratus
    Dec 25 '18 at 1:53






  • 1




    $begingroup$
    Nicely done, for sure ! $to +1$
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 3:25






  • 2




    $begingroup$
    Consider abandoning bigg( and bigg) and replacing them by left( and right).
    $endgroup$
    – Did
    Dec 25 '18 at 9:36


















7












$begingroup$


I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.



Proof:



We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$






sequences-and-series proof-verification fourier-series decimal-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is a great way to obtain some nice problems for math contests or exams :D
    $endgroup$
    – Zacky
    Dec 25 '18 at 1:35






  • 1




    $begingroup$
    Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
    $endgroup$
    – Kemono Chen
    Dec 25 '18 at 1:50






  • 2




    $begingroup$
    @Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
    $endgroup$
    – clathratus
    Dec 25 '18 at 1:53






  • 1




    $begingroup$
    Nicely done, for sure ! $to +1$
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 3:25






  • 2




    $begingroup$
    Consider abandoning bigg( and bigg) and replacing them by left( and right).
    $endgroup$
    – Did
    Dec 25 '18 at 9:36
















7












7








7


2



$begingroup$


I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.



Proof:



We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$






sequences-and-series proof-verification fourier-series decimal-expansion







share|cite|improve this question











$endgroup$




I am looking for numerical evidence that
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinleft(frac{n}{m^2}right)=frac{pi^6}{11340}-frac{pi^4}{72}$$
I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.



Proof:



We know that for $|t|leqpi$,
$$t^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nt$$
Solving for the sum then integrating both sides from $0$ to $x$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sin(nx)=frac{x^3}{12}-frac{pi^2x}{12}$$
Then plugging in $x=frac1{m^2}$ for integer $mgeq1$,
$$sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{1}{12m^6}-frac{pi^2}{12m^2}$$
then applying $sum_{mgeq1}$ on both sides,
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac1{12}zeta(6)-frac{pi^2}{12}zeta(2)$$
We simplify to reach our conclusion
$$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^3}sinbigg(frac{n}{m^2}bigg)=frac{pi^6}{11340}-frac{pi^4}{72}$$






sequences-and-series proof-verification fourier-series decimal-expansion




sequences-and-series proof-verification fourier-series decimal-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 9:35









Did

248k23225463




248k23225463










asked Dec 25 '18 at 1:00









clathratusclathratus

4,9601338




4,9601338








  • 1




    $begingroup$
    This is a great way to obtain some nice problems for math contests or exams :D
    $endgroup$
    – Zacky
    Dec 25 '18 at 1:35






  • 1




    $begingroup$
    Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
    $endgroup$
    – Kemono Chen
    Dec 25 '18 at 1:50






  • 2




    $begingroup$
    @Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
    $endgroup$
    – clathratus
    Dec 25 '18 at 1:53






  • 1




    $begingroup$
    Nicely done, for sure ! $to +1$
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 3:25






  • 2




    $begingroup$
    Consider abandoning bigg( and bigg) and replacing them by left( and right).
    $endgroup$
    – Did
    Dec 25 '18 at 9:36
















  • 1




    $begingroup$
    This is a great way to obtain some nice problems for math contests or exams :D
    $endgroup$
    – Zacky
    Dec 25 '18 at 1:35






  • 1




    $begingroup$
    Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
    $endgroup$
    – Kemono Chen
    Dec 25 '18 at 1:50






  • 2




    $begingroup$
    @Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
    $endgroup$
    – clathratus
    Dec 25 '18 at 1:53






  • 1




    $begingroup$
    Nicely done, for sure ! $to +1$
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 3:25






  • 2




    $begingroup$
    Consider abandoning bigg( and bigg) and replacing them by left( and right).
    $endgroup$
    – Did
    Dec 25 '18 at 9:36










1




1




$begingroup$
This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35




$begingroup$
This is a great way to obtain some nice problems for math contests or exams :D
$endgroup$
– Zacky
Dec 25 '18 at 1:35




1




1




$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50




$begingroup$
Mathematica is able to find the value of the sum up to $5$ decimal places: $-1.26813$. It agrees with the numerical value of $frac{pi^6}{11340}-frac{pi^4}{72}$.
$endgroup$
– Kemono Chen
Dec 25 '18 at 1:50




2




2




$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53




$begingroup$
@Zacky using a similar method, one can show that $$sum_{mgeq1}sum_{ngeq1}frac{(-1)^n}{n^4}cos(n/m^2)=frac{pi^6}{2160}-frac{pi^8}{453600}-frac{7pi^4}{720}$$
$endgroup$
– clathratus
Dec 25 '18 at 1:53




1




1




$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25




$begingroup$
Nicely done, for sure ! $to +1$
$endgroup$
– Claude Leibovici
Dec 25 '18 at 3:25




2




2




$begingroup$
Consider abandoning bigg( and bigg) and replacing them by left( and right).
$endgroup$
– Did
Dec 25 '18 at 9:36






$begingroup$
Consider abandoning bigg( and bigg) and replacing them by left( and right).
$endgroup$
– Did
Dec 25 '18 at 9:36












1 Answer
1






active

oldest

votes


















5












$begingroup$

After you so elegant solution, I am quite ashamed to have done it the following way.



$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$



Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$



It takes a long time to arrive to $-1.268125453640218644419$.



Update



I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$



Edit



I tried to understand why this computation was taking so much computing time.



Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.



This explains that.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Elegant or not, I really appreciate the answer. Thanks Claude :)
    $endgroup$
    – clathratus
    Dec 25 '18 at 4:07






  • 1




    $begingroup$
    @clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 4:12










  • $begingroup$
    Nice! Mind if I ask you what programm do you use for such computations?
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:04










  • $begingroup$
    @Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 15:09










  • $begingroup$
    I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:11













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

After you so elegant solution, I am quite ashamed to have done it the following way.



$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$



Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$



It takes a long time to arrive to $-1.268125453640218644419$.



Update



I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$



Edit



I tried to understand why this computation was taking so much computing time.



Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.



This explains that.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Elegant or not, I really appreciate the answer. Thanks Claude :)
    $endgroup$
    – clathratus
    Dec 25 '18 at 4:07






  • 1




    $begingroup$
    @clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 4:12










  • $begingroup$
    Nice! Mind if I ask you what programm do you use for such computations?
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:04










  • $begingroup$
    @Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 15:09










  • $begingroup$
    I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:11


















5












$begingroup$

After you so elegant solution, I am quite ashamed to have done it the following way.



$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$



Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$



It takes a long time to arrive to $-1.268125453640218644419$.



Update



I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$



Edit



I tried to understand why this computation was taking so much computing time.



Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.



This explains that.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Elegant or not, I really appreciate the answer. Thanks Claude :)
    $endgroup$
    – clathratus
    Dec 25 '18 at 4:07






  • 1




    $begingroup$
    @clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 4:12










  • $begingroup$
    Nice! Mind if I ask you what programm do you use for such computations?
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:04










  • $begingroup$
    @Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 15:09










  • $begingroup$
    I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:11
















5












5








5





$begingroup$

After you so elegant solution, I am quite ashamed to have done it the following way.



$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$



Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$



It takes a long time to arrive to $-1.268125453640218644419$.



Update



I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$



Edit



I tried to understand why this computation was taking so much computing time.



Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.



This explains that.






share|cite|improve this answer











$endgroup$



After you so elegant solution, I am quite ashamed to have done it the following way.



$$sum_{n=1}^inftyfrac{(-1)^n}{n^3}sin(nx)=frac{i}{2} left(text{Li}_3left(-e^{-i x}right)-text{Li}_3left(-e^{i x}right)right)$$



Replace $x=frac 1 {m^2}$ and numerically compute
$$frac{i}{2}sum_{m=1}^infty left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$



It takes a long time to arrive to $-1.268125453640218644419$.



Update



I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer
$$-1.26812545364021864441879478985797582963162172461540$$



Edit



I tried to understand why this computation was taking so much computing time.



Let us consider
$$a_m=frac{i}{2} left(text{Li}_3left(-e^{-frac i {m^2} }right)-text{Li}_3left(-e^{frac i {m^2}}right)right)$$ For large values of $m$, we have
$$a_m=-frac{pi ^2}{12 m^2}+frac{1}{12 m^6}+Oleft(frac{1}{m^{14}}right)$$ So
$$sum_{m=1}^infty a_msimeqsum_{m=1}^p a_m+sum_{m=p+1}^inftyleft(-frac{pi ^2}{12 m^2}+frac{1}{12 m^6} right)=sum_{m=1}^p a_m+frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}$$ and the last term is almost
$$frac{psi ^{(5)}(p+1)-120 pi ^2 psi ^{(1)}(p+1)}{1440}simeq -frac{pi^2}{12p}left(1-frac{1}{2 p}+frac{1}{6 p^2}+Oleft(frac{1}{p^4}right) right)$$ In other words, for $k$ significant figures, we need to add a lot terms.



This explains that.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 25 '18 at 14:39

























answered Dec 25 '18 at 4:03









Claude LeiboviciClaude Leibovici

124k1157135




124k1157135












  • $begingroup$
    Elegant or not, I really appreciate the answer. Thanks Claude :)
    $endgroup$
    – clathratus
    Dec 25 '18 at 4:07






  • 1




    $begingroup$
    @clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 4:12










  • $begingroup$
    Nice! Mind if I ask you what programm do you use for such computations?
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:04










  • $begingroup$
    @Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 15:09










  • $begingroup$
    I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:11




















  • $begingroup$
    Elegant or not, I really appreciate the answer. Thanks Claude :)
    $endgroup$
    – clathratus
    Dec 25 '18 at 4:07






  • 1




    $begingroup$
    @clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 4:12










  • $begingroup$
    Nice! Mind if I ask you what programm do you use for such computations?
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:04










  • $begingroup$
    @Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
    $endgroup$
    – Claude Leibovici
    Dec 25 '18 at 15:09










  • $begingroup$
    I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
    $endgroup$
    – Zacky
    Dec 25 '18 at 15:11


















$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07




$begingroup$
Elegant or not, I really appreciate the answer. Thanks Claude :)
$endgroup$
– clathratus
Dec 25 '18 at 4:07




1




1




$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12




$begingroup$
@clathratus. It is not only elegant; it is beautiful ! Beauty is part of mathematics.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 4:12












$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04




$begingroup$
Nice! Mind if I ask you what programm do you use for such computations?
$endgroup$
– Zacky
Dec 25 '18 at 15:04












$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09




$begingroup$
@Zacky. I have an old CAS that, in my group over more than twenty years, we adapted to our specific needs. I suppose that this could easily be done using Mathematica.
$endgroup$
– Claude Leibovici
Dec 25 '18 at 15:09












$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11






$begingroup$
I see, pretty nice, I only have the online version for Mathematica. If you have some free time, can you please check it's power against this? $$int_0^infty frac{1-cos x}{8-4xsin x +x^2(1-cos x)}dx=frac{pi}{4}$$ My wolfram can't even spit $1$ decimal out... So I would be glad if I know that the result it's correct. It is $I_5$ that I put here: math.stackexchange.com/a/3027759/515527
$endgroup$
– Zacky
Dec 25 '18 at 15:11




















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