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Hello.

I have been trying very hard to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$ and could not quite get anywhere. This inequality has been verified by computer for $k\le40$.

Some clues that might work (kindly provided by Doron Zeilberger) are as follows:

  1. Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}((-1)^{(j+1)}*binomial(2*k+1,j)*E^j$P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

  2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2*P_{k-1}(E)=Some Binomial In E$;

  3. The original sum can be expressed as $P_k(E)x^{(2*k-2)} | x=k $;

  4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$ before plugging-in $x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.

Unfortunately I do not know how to find a recurrence such as suggested in 4.

I would appreciate any help that members of the MathOverflow community can provide.
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Need help proving that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} binomial{2k+1,j} \binom{2k+1}{j} \ge 0$

Hello.

I have been trying very hard to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} binomial(2k+1,j) \binom{2k+1}{j} \ge 0$ and could not quite get anywhere. This inequality has been verified by computer for $k\le40$.

Some clues that might work (kindly provided by Doron Zeilberger) are as follows:

  1. Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}((-1)^{(j+1)}*binomial(2*k+1,j)*E^j$;

  2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2*P_{k-1}(E)=Some Binomial In E$;

  3. The original sum can be expressed as $P_k(E)x^{(2*k-2)} | x=k $;

  4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$ before plugging-in $x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.

Unfortunately I do not know how to find a recurrence such as suggested in 4.

I would appreciate any help that members of the MathOverflow community can provide.
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Need help proving that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} binomial{2k+1,j} \ge 0$

Hello.

I have been trying very hard to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} binomial(2k+1,j) \ge 0$ and could not quite get anywhere. This inequality has been verified by computer for $k\le40$.

Some clues that might work (kindly provided by Doron Zeilberger) are as follows:

  1. Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}((-1)^{(j+1)}*binomial(2*k+1,j)*E^j$;

  2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2*P_{k-1}(E)=Some Binomial In E$;

  3. The original sum can be expressed as $P_k(E)x^{(2*k-2)} | x=k $;

  4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$ before plugging-in $x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.

Unfortunately I do not know how to find a recurrence such as suggested in 4.

I would appreciate any help that members of the MathOverflow community can provide.