Skip to main content
Mathematics

Return to Question

added 1 character in body
Source Link
Ali Olaikhan
Ali Olaikhan
  • 28.2k
  • 4
  • 68
  • 150

How to prove that

$$\sum_{n=1}^\infty\frac{H_n^4}{n^2}=\frac{979}{24}\zeta(6)+3\zeta^2(3)\ ?$$ $$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}=\frac{41}{12}\zeta(6)+2\zeta^2(3)\ ?$$ where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

We can find these two sums nicely evaluated in the book Almost Impossible Integrals, Sums and series page 429 using pure series manipulations and I managed to find their closed forms using integration and some harmonic-number identities ( Solution will be posted in the answer section ), but can we tackle them by other means like contour integration or the $\arcsin^4x$ identity just like what @nospoon did in his solution here, where he mentioned the result of the second sum in $(4)$ but not the first one, or by any other way?

Thanks  .

How to prove that

$$\sum_{n=1}^\infty\frac{H_n^4}{n^2}=\frac{979}{24}\zeta(6)+3\zeta^2(3)\ ?$$ $$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}=\frac{41}{12}\zeta(6)+2\zeta^2(3)\ ?$$ where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

We can find these two sums nicely evaluated in the book Almost Impossible Integrals, Sums and series page 429 using pure series manipulations and I managed to find their closed forms using integration and some harmonic-number identities ( Solution will be posted in the answer section ), but can we tackle them by other means like contour integration or the $\arcsin^4x$ identity just like what @nospoon did in his solution here, where he mentioned the result of the second sum in $(4)$ but not the first one, or by any other way?

Thanks.

How to prove that

$$\sum_{n=1}^\infty\frac{H_n^4}{n^2}=\frac{979}{24}\zeta(6)+3\zeta^2(3)\ ?$$ $$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}=\frac{41}{12}\zeta(6)+2\zeta^2(3)\ ?$$ where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

We can find these two sums nicely evaluated in the book Almost Impossible Integrals, Sums and series page 429 using pure series manipulations and I managed to find their closed forms using integration and some harmonic-number identities ( Solution will be posted in the answer section ), but can we tackle them by other means like contour integration or the $\arcsin^4x$ identity just like what @nospoon did in his solution here, where he mentioned the result of the second sum in $(4)$ but not the first one, or by any other way?

Thanks  .

removed (imo) redundant part of title
Link
ViktorStein
ViktorStein
  • 5.1k
  • 6
  • 23
  • 60

Two advanced sums: Compute $\sum_{n=1}^\infty\frac{H_n^4}{n^2}$ and $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}$

Source Link
Ali Olaikhan
Ali Olaikhan
  • 28.2k
  • 4
  • 68
  • 150

Two advanced sums: Compute $\sum_{n=1}^\infty\frac{H_n^4}{n^2}$ and $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}$

How to prove that

$$\sum_{n=1}^\infty\frac{H_n^4}{n^2}=\frac{979}{24}\zeta(6)+3\zeta^2(3)\ ?$$ $$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}=\frac{41}{12}\zeta(6)+2\zeta^2(3)\ ?$$ where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

We can find these two sums nicely evaluated in the book Almost Impossible Integrals, Sums and series page 429 using pure series manipulations and I managed to find their closed forms using integration and some harmonic-number identities ( Solution will be posted in the answer section ), but can we tackle them by other means like contour integration or the $\arcsin^4x$ identity just like what @nospoon did in his solution here, where he mentioned the result of the second sum in $(4)$ but not the first one, or by any other way?

Thanks.