English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

In this paper, via the beta function we evaluate some series of the type$\sum_{k=0}^\infty(ak+b)x^k/\binom{mk}{nk}$. We completely determine the valuesof $$\sum_{k=1}^\infty\frac {k^rx^k}{\binom{3k}k}\\left(-\frac{27}4<x<\frac{27}4\right)\ \text{and}\ \sum_{k=1}^\infty\frac{k^rx^k}{\binom{4k}{2k}}\ (-16<x<16)$$ for $r=0,\pm1$. For example, we provethat $$\sum_{k=0}^\infty\frac{(49k+1)8^k}{3^k\binom{3k}k}=81+16\sqrt3\,\pi \ \\text{and}\ \ \sum_{k=0}^\infty\frac{10k-1}{\binom{4k}{2k}}=\frac{4\sqrt3}{27}\pi.$$ 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\logn-32n(n+1)^2(n^2-4n+1), \end{align*} where$$P(n):=n^6-58n^5+159n^4+52n^3+159n^2-58n+1.$$

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