1
$\begingroup$

Let $$D_{n}=\sum_{j=0}^{n-1}(-1)^{n+j-1}\dfrac{\binom{2n-4}{j}}{n+j-1},E_{n}=\sum_{j=0}^{n-1}(-1)^{n+j-1}\dfrac{\binom{2n-4}{j}}{n+j}$$

I want find $D_{n}$ and $E_{n}$ recursive relations,

I try take somedays,At present I only find following $$\sum_{j=0}^{2n-4}(-1)^j\dfrac{\binom{2n-4}{j}}{n+j-1}=\dfrac{3n-4}{n-1}\sum_{j=0}^{2n-4}(-1)^j\dfrac{\binom{2n-4}{j}}{n+j}$$ can you someone help find the recursive relation with $D_{n}$ and $E_{n}$?

I also post MO

$\endgroup$
1
$\begingroup$

The series $D_{n}$ and $E_{n}$ are of the form \begin{align} D_{n} &= \sum_{j=0}^{n-1} (-1)^{n+j-1} \frac{\binom{2n-4}{j}}{n+j-1} \\ &= \frac{ \binom{2n-4}{n} }{ (2n-1) \Gamma(3n-4) } \, {}_{3}F_{2}(1, 4-n, 2n-1; n+1, 2n; 1) - (-1)^{n} \, \frac{\Gamma(n) \Gamma(2n-3)}{(n-1) \Gamma(3n-4)} \end{align} and \begin{align} E_{n} &= \sum_{j=0}^{n-1} (-1)^{n+j-1} \frac{\binom{2n-4}{j}}{n+j} \\ &= \frac{1}{2n} \, \binom{2n-4}{n} \, {}_{3}F_{2}( 1, 4-n, 2n; n+1, 2n+1; 1) - (-1)^{n} \, \frac{\Gamma(n) \Gamma(2n-3)}{\Gamma(3n-3)}. \end{align} By using the recurrence relations from Wolfram generalized hypergeometric Functions page an attempt can be made to make a recurrence relation for both $D_{n}$ and $E_{n}$.

$\endgroup$
  • $\begingroup$ Thank you,But we also have relation? $\endgroup$ – math110 Mar 19 '15 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.