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Marco Golla
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I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. 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)}*\binom{2*k+1}{j}*E^j$$P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}\binom{2k+1}{j}E^j$;

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

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

  4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$$P_k(E)x^{(2k-2)}$ before plugging-in 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 by clue 4.

I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. 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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

  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 by clue 4.

I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. 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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}\binom{2k+1}{j}E^j$;

  2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2P_{k-1}(E) =$ some binomial in $E$;

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

  4. Try to derive a recurrence for $P_k(E)x^{(2k-2)}$ 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 by clue 4.

Need help proving that Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$0?$

I am trying to show thatprove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. 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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

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

  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 to facilitate the induction.

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

Need help proving that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$

I am trying to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. 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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

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

  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.

Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$

I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. 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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

  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 by clue 4.

Hello.

I have beenam 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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

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

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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

  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.

I am trying to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. 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 $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

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

  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.

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Gjergji Zaimi
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