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Zhi-Wei Sun
Zhi-Wei Sun
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Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}$$ and that $100+299k = 100(1+3k)-k$ and $24k^2+8k = 8k(3k+1)$,. soSo $$\frac{100+299k}{24k^2+8k}=\frac{100}{8k}-\frac1{8(3k+1)}.$$ SoTherefore the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083, and Batir's papers cited there. For $|x|<27/4$, we also have $$\sum_{k=1}^n((4x-27)k+2x+27)\frac{x^k}{\binom{3k}k}-6\sum_{k=1}^n\frac{x^k}{k\binom{3k}k}=-2x+\frac{(4n+2)x^{n+1}}{\binom{3n}n}$$ which tends to $-2x$ as $n\to\infty$.

Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}$$ and that $100+299k = 100(1+3k)-k$ and $24k^2+8k = 8k(3k+1)$, so $$\frac{100+299k}{24k^2+8k}=\frac{100}{8k}-\frac1{8(3k+1)}.$$ So the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083 and Batir's papers cited there.

Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}$$ and that $100+299k = 100(1+3k)-k$. So $$\frac{100+299k}{24k^2+8k}=\frac{100}{8k}-\frac1{8(3k+1)}.$$ Therefore the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083, and Batir's papers cited there. For $|x|<27/4$, we also have $$\sum_{k=1}^n((4x-27)k+2x+27)\frac{x^k}{\binom{3k}k}-6\sum_{k=1}^n\frac{x^k}{k\binom{3k}k}=-2x+\frac{(4n+2)x^{n+1}}{\binom{3n}n}$$ which tends to $-2x$ as $n\to\infty$.

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Christophe Leuridan
Christophe Leuridan
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Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}.$$ So$$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}$$ and that $100+299k = 100(1+3k)-k$ and $24k^2+8k = 8k(3k+1)$, so $$\frac{100+299k}{24k^2+8k}=\frac{100}{8k}-\frac1{8(3k+1)}.$$ So the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083 and Batir's papers cited there.

Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}.$$ So the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083 and Batir's papers cited there.

Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}$$ and that $100+299k = 100(1+3k)-k$ and $24k^2+8k = 8k(3k+1)$, so $$\frac{100+299k}{24k^2+8k}=\frac{100}{8k}-\frac1{8(3k+1)}.$$ So the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083 and Batir's papers cited there.

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Zhi-Wei Sun
Zhi-Wei Sun
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Note that $$\frac1{3k(3k+1)}=\frac1{3k}-\frac1{3k+1}.$$ So the problem reduces to the evaluations of the sums $$\sum_{k=1}^\infty\frac{x^k}{k^r\binom{3k}k}\ \text{and}\ \sum_{k=0}^\infty \frac{x^k}{(3k+1)\binom{3k}k}$$ with $x=1/576$ and $r\in\{0,\pm1\}$. These sums with $|x|<27/4$ have been evaluated, see this paper, this paper, arXiv:2401.12083 and Batir's papers cited there.