Sum $\sum\limits_{j=k}^n\binom{j}{k}(-1)^j$

Question

I am looking for a closed form of this sum:$\sum\limits_{j=k}^n\binom{j}{k}(-1)^j$

I know that this sum has a closed form:

$\sum\limits_{j=k}^n\binom{j}{k}=\binom{n+1}{k+1}$

I can get this closed form if I have the following in closed form:

$\sum\limits_{j=0}^n\binom{2j}{k}$

Answer 1

In fact, your sum turns out into $$ \begin{array}{l} \sum\limits_{\left( {0\, \le \,m\, \le } \right)\,j\, \le \,n} {\left( \begin{array}{c} j \\ m \\ \end{array} \right)\left( { - 1} \right)^{\,j} } = \left( { - 1} \right)^{\,m} \sum\limits_{\left( {0\, \le \,m\, \le } \right)\,j\, \le \,n} {\left( \begin{array}{c} j \\ j - m \\ \end{array} \right)\left( { - 1} \right)^{\,j - m} } = \\ = \left( { - 1} \right)^{\,m} \sum\limits_{\left( {0\, \le \,} \right)\,k\, \le \,n - m} {\left( \begin{array}{c} - m - 1 \\ k \\ \end{array} \right)} \\ \end{array} $$ and no simple formula is known for the trunked binomial expansion, if not in terms of Hypergeometric function as already commented.
However, it may be of interest to know that it is possible to express a compact formal power series for it (z-transform): $$ \begin{gathered} F(x,z) = \sum\limits_{\,0\, \leqslant \,k} {\left( {\sum\limits_{\left( {0\, \leqslant \,m\, \leqslant } \right)\,j\, \leqslant \,k} {\left( \begin{gathered} j \\ m \\ \end{gathered} \right)x^{\,j} } } \right)z^{\,k} } = \hfill \\ = \sum\limits_{\,0\, \leqslant \,k} {\left( {\sum\limits_{0\, \leqslant \;j} {\left( \begin{gathered} k - j \\ k - j \\ \end{gathered} \right)\left( \begin{gathered} j \\ m \\ \end{gathered} \right)x^{\,j} } } \right)z^{\,k} } = \hfill \\ = \sum\limits_{\,\left\{ \begin{subarray}{l} 0\, \leqslant \,k \\ 0\, \leqslant \;j \end{subarray} \right.} {\left( \begin{gathered} k - j \\ k - j \\ \end{gathered} \right)\left( \begin{gathered} j \\ m \\ \end{gathered} \right)\left( {xz} \right)^{\,j} z^{\,k - j} } = \hfill \\ = \frac{1} {{1 - z}}\sum\limits_{0\, \leqslant \;j} {\left( \begin{gathered} j \\ m \\ \end{gathered} \right)\left( {xz} \right)^{\,j} } = \frac{{x^{\,m} z^{\,m} }} {{\left( {1 - z} \right)\left( {1 - xz} \right)^{\,m + 1} }} \hfill \\ \end{gathered} $$ so, in your case: $$ F( - 1,z) = \sum\limits_{\,0\, \le \,k} {\left( {\sum\limits_{\left( {0\, \le \,m\, \le } \right)\,j\, \le \,k} {\left( \begin{array}{c} j \\ m \\ \end{array} \right)(-1)^{\,j} } } \right)z^{\,k} } = \frac{{\left( { - 1} \right)^{\,m} z^{\,m} }}{{\left( {1 - z} \right)\left( {1 + z} \right)^{\,m + 1} }} $$

Answer 2

This is table https://oeis.org/A239473 in the OEIS (apart from a sign flip in each second row). Because there is on closed form mentioned there, I guess there is none.