论文标题
评估某些类型的$ \ sum_ {k = 0}^\ infty（ak+b）x^k/\ binom {mk} {nk} $
Evaluations of some series of the type $\sum_{k=0}^\infty(ak+b)x^k/\binom{mk}{nk}$
论文作者
论文摘要
在本文中，通过beta函数，我们评估了一些类型的$ \ sum_ {k = 0}^\ infty（ak+b）x^k/\ binom {mk} {nk} $。例如，我们证明$ \ sum_ {k = 0}^\ infty \ frac {（49k+1）8^k} {3^k \ binom {3K} k} = 81+16 \ sqrt3 \ sqrt3 \， \ sum_ {k = 0}^\ infty \ frac {10k-1} {\ binom {4K} {2k}} = \ frac {4 \ sqrt 3} {27} {27}π$$ 我们还建立了以下有效公式，用于计算$ \ log n $，$ 1 <n \ le 85/4 $：\ begin {align*}＆\ sum_ {k = 0}^\ infty \ frac {（2（n^2+6n+1）^2（n^2+6n+1）^2（n^2-10n+1） {（-n）^k（n+1）^{2K} \ binom {4K} {2K} {2K}}}}} \ \ \ \ \ \ \ \ \＆= 6n（n+1）（n-1）（n-1）^3 \ log n-32n（n+1） $$ p（n）：= n^6-58n^5+159n^4+52n^3+159n^2-58n+1。 （k \ in \ mathbb n）$。
In this paper, via the beta function we evaluate some series of the type $\sum_{k=0}^\infty(ak+b)x^k/\binom{mk}{nk}$. For example, we prove that $$\sum_{k=0}^\infty\frac{(49k+1)8^k}{3^k\binom{3k}k}=81+16\sqrt3\,π\ \ \text{and}\ \ \sum_{k=0}^\infty\frac{10k-1}{\binom{4k}{2k}}=\frac{4\sqrt 3}{27}π.$$ We also establish the following efficient formula for computing $\log n$ with $1<n\le 85/4$: \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 $$P(n):=n^6-58n^5+159n^4+52n^3+159n^2-58n+1.$$ In addition, we pose some conjectures on series whose summands involve $\binom{2k}k/(\binom{3k}k\binom{6k}{3k})\ (k\in\mathbb N)$.
