Timeline for Is it true that $\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{(2k+1)16^k}$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 20, 2020 at 11:35 | vote | accept | Zhi-Wei Sun | ||
| May 28, 2019 at 0:22 | answer | added | Nemo | timeline score: 12 | |
| Feb 15, 2019 at 22:12 | comment | added | Timothy Budd | See also J. Campbell, "New series involving harmonic numbers and squared central binomial coefficients" (2018), hal.archives-ouvertes.fr/hal-01774708, around equation (1.6) for some remarks on the conjecture. | |
| Feb 15, 2019 at 18:31 | comment | added | Henri Cohen | To 38D, in 1 line of Pari/GP both sides are equal to 0.24365784900593188639480303875745624291 | |
| Feb 15, 2019 at 14:42 | comment | added | Nemo | @Zhi-WeiSun it is possible to convert these sums into integrals of elliptic functions using formulas mentioned in the comment above. I think one can verify your formula to at least several hundred digits by this approach. | |
| Feb 15, 2019 at 13:53 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
added 42 characters in body; edited title
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| Feb 15, 2019 at 13:21 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
Add motivation.
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| Feb 15, 2019 at 12:40 | comment | added | Zhi-Wei Sun | I have heared that Don Zagier has a method to accelerate a slowly convergent series. Is anybody here familiar with the method? | |
| Feb 15, 2019 at 12:05 | comment | added | Wolfgang | If I have done it correctly, you can restate it as $$2+\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(3H_k-2H_{2k})= \sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{k(2k+1)16^k} $$with better numerical convergence. | |
| Feb 15, 2019 at 10:58 | comment | added | Carlo Beenakker | what is the basis for the conjectured equality? a numerical evaluation of the two sides of the equation gives a 1% difference with an error estimate of $10^{-3}$... | |
| Feb 15, 2019 at 9:17 | comment | added | Nemo | Probably this might help $$\sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_n}{16^n}k^{2n}=K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k^2}{16(1-k^2)},\tag{1}$$ $$\sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_{2n}}{16^n}k^{2n}=\frac12K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k}{4(1-k^2)}.\tag{2}$$ | |
| Feb 15, 2019 at 9:07 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |