Finite sum of integers inverses - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T12:29:09Z http://mathoverflow.net/feeds/question/67784 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67784/finite-sum-of-integers-inverses Finite sum of integers inverses ivane 2011-06-14T17:53:22Z 2011-06-15T17:37:01Z <p>Recently, we learned from Renzo Sprugnoli that $\sum_{k=0}^n \frac{4^k}{B_k} =\frac{2n+1}{3}\frac{4^{n+1}}{B_{n+1}}+\frac{1}{3}$, where $B_n$ are the famous <strong>central binomial coefficients</strong>, $B_n={2n \choose n}$.</p> <p>Nowadays it is posible to find; using W|A, <em>The Wolframalpha Calculator</em>, that:</p> <p>$$\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{2n^2+5n+2}{5}\frac{4^{n+1}}{B_{n+1}}+\frac{1}{5}.$$</p> <p>My "begs" are for someone to help me to pursuit some interesant generalizations or applications about the involved <strong>Catalan numbers</strong>. </p> <p><strong>Ref:</strong> R. Sprugnoli, "Sum of reciprocals of the central binomial coefficients", INTEGERS: Electronic journal of combinatorial number theory 6 (2006), #A27. </p> <p><strong>Update:</strong></p> <p>Is the formula $\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{2n^2+5n+2}{5}\frac{4^{n+1}}{B_{n+1}}+\frac{1}{5}$ well known?</p>