Evaluations of some series of the type $\sum_{k=0}^\infty(ak+b)x^k/\binom{mk}{nk}$

In this paper, via the beta function we evaluate some series of the type $\sum_{k = 0}^{\infty} (a k + b) x^{k} / (\binom{m k}{n k})$ . For example, we prove that $\sum_{k = 0}^{\infty} \frac{(49 k + 1) 8^{k}}{3^{k} (\binom{3 k}{k})} = 81 + 16 \sqrt{3} π and \sum_{k = 0}^{\infty} \frac{10 k - 1}{(\binom{4 k}{2 k})} = \frac{4 \sqrt{3}}{27} π .$ We also establish the following efficient formula for computing $\log n$ with $1 < n \leq 85 / 4$ : \begin{align*} &\sum_{k=0}^\infty\frac{(2(n^2+6n+1)^2(n^2-10n+1)k+P(n))(n-1)^{4k}} {(-n)^k(n+1)^{2k}\binom{4k}{2k}}\ \ \ &=6n(n+1)(n-1)^3\log n-32n(n+1)^2(n^2-4n+1), \end{align*} where $P (n) : = n^{6} - 58 n^{5} + 159 n^{4} + 52 n^{3} + 159 n^{2} - 58 n + 1.$ In addition, we pose some conjectures on series whose summands involve $(\binom{2 k}{k}) / ((\binom{3 k}{k}) (\binom{6 k}{3 k})) (k \in N)$ .

Evaluations of some series of the type $\sum_{k = 0}^{\infty} (a k + b) x^{k} / (\binom{m k}{n k})$

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Evaluations of some series of the type $\sum_{k = 0}^{\infty} (a k + b) x^{k} / (\binom{m k}{n k})$

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Evaluations of some series of the type ∑k=0∞(ak+b)xk/(mknk)∑k=0∞​(ak+b)xk/(nkmk​)

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Evaluations of some series of the type ∑k=0∞(ak+b)xk/(mknk)∑k=0∞​(ak+b)xk/(nkmk​)

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Evaluations of some series of the type $\sum_{k = 0}^{\infty} (a k + b) x^{k} / (\binom{m k}{n k})$

讲解博客

AI智能校验

Evaluations of some series of the type $\sum_{k = 0}^{\infty} (a k + b) x^{k} / (\binom{m k}{n k})$

讲解博客

AI智能校验