Evaluation of Euler-type sum $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}$ [closed]












-1












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How can one evaluate the sum $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}$? Here $H_{n}$ denotes $n$-th harmonic number.






real-analysis sequences-and-series summation euler-sums







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closed as off-topic by amWhy, Masacroso, Clarinetist, RRL, user10354138 Dec 7 '18 at 2:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Masacroso, Clarinetist, RRL, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    What sum will come next?
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:57










  • $begingroup$
    with a software and a computer? Wolfram Mathematica gives the approximate result of $0.880734$
    $endgroup$
    – Masacroso
    Dec 6 '18 at 23:18










  • $begingroup$
    $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}=8Li_4(frac{1}{2})+frac{pi^2}{6}ln^22+frac{1}{3}ln^42-frac{61}{1440}{pi^4}$
    $endgroup$
    – user178256
    Dec 7 '18 at 8:49










  • $begingroup$
    How can this be proven?
    $endgroup$
    – Isak
    Dec 7 '18 at 10:23
















-1












$begingroup$


How can one evaluate the sum $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}$? Here $H_{n}$ denotes $n$-th harmonic number.






real-analysis sequences-and-series summation euler-sums







share|cite|improve this question









$endgroup$



closed as off-topic by amWhy, Masacroso, Clarinetist, RRL, user10354138 Dec 7 '18 at 2:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Masacroso, Clarinetist, RRL, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    What sum will come next?
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:57










  • $begingroup$
    with a software and a computer? Wolfram Mathematica gives the approximate result of $0.880734$
    $endgroup$
    – Masacroso
    Dec 6 '18 at 23:18










  • $begingroup$
    $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}=8Li_4(frac{1}{2})+frac{pi^2}{6}ln^22+frac{1}{3}ln^42-frac{61}{1440}{pi^4}$
    $endgroup$
    – user178256
    Dec 7 '18 at 8:49










  • $begingroup$
    How can this be proven?
    $endgroup$
    – Isak
    Dec 7 '18 at 10:23














-1












-1








-1


1



$begingroup$


How can one evaluate the sum $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}$? Here $H_{n}$ denotes $n$-th harmonic number.






real-analysis sequences-and-series summation euler-sums







share|cite|improve this question









$endgroup$




How can one evaluate the sum $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}$? Here $H_{n}$ denotes $n$-th harmonic number.






real-analysis sequences-and-series summation euler-sums




real-analysis sequences-and-series summation euler-sums






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 22:52









IsakIsak

215




215




closed as off-topic by amWhy, Masacroso, Clarinetist, RRL, user10354138 Dec 7 '18 at 2:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Masacroso, Clarinetist, RRL, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Masacroso, Clarinetist, RRL, user10354138 Dec 7 '18 at 2:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Masacroso, Clarinetist, RRL, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    What sum will come next?
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:57










  • $begingroup$
    with a software and a computer? Wolfram Mathematica gives the approximate result of $0.880734$
    $endgroup$
    – Masacroso
    Dec 6 '18 at 23:18










  • $begingroup$
    $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}=8Li_4(frac{1}{2})+frac{pi^2}{6}ln^22+frac{1}{3}ln^42-frac{61}{1440}{pi^4}$
    $endgroup$
    – user178256
    Dec 7 '18 at 8:49










  • $begingroup$
    How can this be proven?
    $endgroup$
    – Isak
    Dec 7 '18 at 10:23














  • 1




    $begingroup$
    What sum will come next?
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:57










  • $begingroup$
    with a software and a computer? Wolfram Mathematica gives the approximate result of $0.880734$
    $endgroup$
    – Masacroso
    Dec 6 '18 at 23:18










  • $begingroup$
    $displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}=8Li_4(frac{1}{2})+frac{pi^2}{6}ln^22+frac{1}{3}ln^42-frac{61}{1440}{pi^4}$
    $endgroup$
    – user178256
    Dec 7 '18 at 8:49










  • $begingroup$
    How can this be proven?
    $endgroup$
    – Isak
    Dec 7 '18 at 10:23








1




1




$begingroup$
What sum will come next?
$endgroup$
– gammatester
Dec 6 '18 at 22:57




$begingroup$
What sum will come next?
$endgroup$
– gammatester
Dec 6 '18 at 22:57












$begingroup$
with a software and a computer? Wolfram Mathematica gives the approximate result of $0.880734$
$endgroup$
– Masacroso
Dec 6 '18 at 23:18




$begingroup$
with a software and a computer? Wolfram Mathematica gives the approximate result of $0.880734$
$endgroup$
– Masacroso
Dec 6 '18 at 23:18












$begingroup$
$displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}=8Li_4(frac{1}{2})+frac{pi^2}{6}ln^22+frac{1}{3}ln^42-frac{61}{1440}{pi^4}$
$endgroup$
– user178256
Dec 7 '18 at 8:49




$begingroup$
$displaystylesum_{n=1}^{infty}frac{H_{n}^{2}}{(2n+1)^{2}}=8Li_4(frac{1}{2})+frac{pi^2}{6}ln^22+frac{1}{3}ln^42-frac{61}{1440}{pi^4}$
$endgroup$
– user178256
Dec 7 '18 at 8:49












$begingroup$
How can this be proven?
$endgroup$
– Isak
Dec 7 '18 at 10:23




$begingroup$
How can this be proven?
$endgroup$
– Isak
Dec 7 '18 at 10:23










1 Answer
1






active

oldest

votes


















2












$begingroup$

$$sum_{ngeq 1}H_n^2 x^n = frac{log^2(1-x)+text{Li}_2(x)}{1-x}$$
hence the explicit computation of your series is equivalent to the computation of
$$ int_{0}^{1}frac{log^2(1-x^2)+text{Li}_2(x^2)}{1-x^2}log(x),dx $$
which can probably be tackled through Fourier(-Legendre) series. The simple part is
$$ int_{0}^{1}frac{log^2(1-x^2)log(x)}{1-x^2},dx = -frac{pi^4}{24}-frac{pi^2}{2}log^2(2)+7zeta(3)log(2).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$?
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:33












  • $begingroup$
    Thank You, that is something yet. Any ideas about the second part?
    $endgroup$
    – Isak
    Dec 6 '18 at 23:33










  • $begingroup$
    @BadamBaplan: it is summing the right thing, since $int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:36










  • $begingroup$
    nice, thanks for explaining. this is a nice approach.
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:38










  • $begingroup$
    @Isak: the difficult part is essentially $sum_{ngeq 1}frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:39


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$$sum_{ngeq 1}H_n^2 x^n = frac{log^2(1-x)+text{Li}_2(x)}{1-x}$$
hence the explicit computation of your series is equivalent to the computation of
$$ int_{0}^{1}frac{log^2(1-x^2)+text{Li}_2(x^2)}{1-x^2}log(x),dx $$
which can probably be tackled through Fourier(-Legendre) series. The simple part is
$$ int_{0}^{1}frac{log^2(1-x^2)log(x)}{1-x^2},dx = -frac{pi^4}{24}-frac{pi^2}{2}log^2(2)+7zeta(3)log(2).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$?
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:33












  • $begingroup$
    Thank You, that is something yet. Any ideas about the second part?
    $endgroup$
    – Isak
    Dec 6 '18 at 23:33










  • $begingroup$
    @BadamBaplan: it is summing the right thing, since $int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:36










  • $begingroup$
    nice, thanks for explaining. this is a nice approach.
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:38










  • $begingroup$
    @Isak: the difficult part is essentially $sum_{ngeq 1}frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:39
















2












$begingroup$

$$sum_{ngeq 1}H_n^2 x^n = frac{log^2(1-x)+text{Li}_2(x)}{1-x}$$
hence the explicit computation of your series is equivalent to the computation of
$$ int_{0}^{1}frac{log^2(1-x^2)+text{Li}_2(x^2)}{1-x^2}log(x),dx $$
which can probably be tackled through Fourier(-Legendre) series. The simple part is
$$ int_{0}^{1}frac{log^2(1-x^2)log(x)}{1-x^2},dx = -frac{pi^4}{24}-frac{pi^2}{2}log^2(2)+7zeta(3)log(2).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$?
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:33












  • $begingroup$
    Thank You, that is something yet. Any ideas about the second part?
    $endgroup$
    – Isak
    Dec 6 '18 at 23:33










  • $begingroup$
    @BadamBaplan: it is summing the right thing, since $int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:36










  • $begingroup$
    nice, thanks for explaining. this is a nice approach.
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:38










  • $begingroup$
    @Isak: the difficult part is essentially $sum_{ngeq 1}frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:39














2












2








2





$begingroup$

$$sum_{ngeq 1}H_n^2 x^n = frac{log^2(1-x)+text{Li}_2(x)}{1-x}$$
hence the explicit computation of your series is equivalent to the computation of
$$ int_{0}^{1}frac{log^2(1-x^2)+text{Li}_2(x^2)}{1-x^2}log(x),dx $$
which can probably be tackled through Fourier(-Legendre) series. The simple part is
$$ int_{0}^{1}frac{log^2(1-x^2)log(x)}{1-x^2},dx = -frac{pi^4}{24}-frac{pi^2}{2}log^2(2)+7zeta(3)log(2).$$






share|cite|improve this answer









$endgroup$



$$sum_{ngeq 1}H_n^2 x^n = frac{log^2(1-x)+text{Li}_2(x)}{1-x}$$
hence the explicit computation of your series is equivalent to the computation of
$$ int_{0}^{1}frac{log^2(1-x^2)+text{Li}_2(x^2)}{1-x^2}log(x),dx $$
which can probably be tackled through Fourier(-Legendre) series. The simple part is
$$ int_{0}^{1}frac{log^2(1-x^2)log(x)}{1-x^2},dx = -frac{pi^4}{24}-frac{pi^2}{2}log^2(2)+7zeta(3)log(2).$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 23:31









Jack D'AurizioJack D'Aurizio

288k33280659




288k33280659












  • $begingroup$
    Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$?
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:33












  • $begingroup$
    Thank You, that is something yet. Any ideas about the second part?
    $endgroup$
    – Isak
    Dec 6 '18 at 23:33










  • $begingroup$
    @BadamBaplan: it is summing the right thing, since $int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:36










  • $begingroup$
    nice, thanks for explaining. this is a nice approach.
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:38










  • $begingroup$
    @Isak: the difficult part is essentially $sum_{ngeq 1}frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:39


















  • $begingroup$
    Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$?
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:33












  • $begingroup$
    Thank You, that is something yet. Any ideas about the second part?
    $endgroup$
    – Isak
    Dec 6 '18 at 23:33










  • $begingroup$
    @BadamBaplan: it is summing the right thing, since $int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:36










  • $begingroup$
    nice, thanks for explaining. this is a nice approach.
    $endgroup$
    – Badam Baplan
    Dec 6 '18 at 23:38










  • $begingroup$
    @Isak: the difficult part is essentially $sum_{ngeq 1}frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 23:39
















$begingroup$
Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$?
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:33






$begingroup$
Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$?
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:33














$begingroup$
Thank You, that is something yet. Any ideas about the second part?
$endgroup$
– Isak
Dec 6 '18 at 23:33




$begingroup$
Thank You, that is something yet. Any ideas about the second part?
$endgroup$
– Isak
Dec 6 '18 at 23:33












$begingroup$
@BadamBaplan: it is summing the right thing, since $int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 23:36




$begingroup$
@BadamBaplan: it is summing the right thing, since $int_{0}^{1}x^{2n}log(x),dx = -frac{1}{(2n+1)^2}$.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 23:36












$begingroup$
nice, thanks for explaining. this is a nice approach.
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:38




$begingroup$
nice, thanks for explaining. this is a nice approach.
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:38












$begingroup$
@Isak: the difficult part is essentially $sum_{ngeq 1}frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 23:39




$begingroup$
@Isak: the difficult part is essentially $sum_{ngeq 1}frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 23:39



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