I know that for a summation such as $$p=\sum_{n=o}^{k}{\binom{k}{n}}=2^k,$$ Wolfram Alpha and some other computational software can easily tell that indeed the above summation evaluates to $2^k$.
I have tried doing a similar thing with summations involving the unsigned Stirling numbers of the first kind ${k \brack n}$, but Wolfram does not simplify such sums. For example, it is well known that the summation $$\sum_{n=0}^{k}{{k \brack n}}$$ evaluates to $k!$, but Wolfram does not simplify the above sum to $k!$. It just leaves the answer as $\sum_{n=0}^{k}{{k \brack n}}$.
I am interested in knowing if any software can simplify sums involving Stirling numbers of the first kind without one specifying lower and upper summation indices. For example, I have been trying to evaluate the sum $$\sum_{n=0}^{l}{{n+v \brack n+2}},$$ but have failed infinitely many times. I was wondering if there is a software that can save me from this stress. Thank you a lot, family!
Unfortunately the sum of Stirling Numbers, either 1st and 2nd kind, does not have a "closed" form, shorter than performing the sum directly.
However for the "diagonal" sum might be interesting an identity which can be derived by the expression through the Eulerian Numbers of 2nd kind $$ \eqalign{ & \left[ \matrix{ x \cr x - n \cr} \right] = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{ n \cr k \cr} \right\rangle } \right\rangle \left( \matrix{ x + k \cr 2n \cr} \right)} = \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{ n \cr k \cr} \right\rangle } \right\rangle \left( \matrix{ k \cr 2n - j \cr} \right)\left( \matrix{ x \cr j \cr} \right)} } \cr} $$ where we use $x$ because this nice identity can be used to extend the definition of the Stirling Numbers also to real and complex values of $x$.
Then summing on $x$ $$ \eqalign{ & \sum\limits_{x = 0}^b {\left[ \matrix{ x \cr x - n \cr} \right]} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{ n \cr k \cr} \right\rangle } \right\rangle \left( \matrix{ k \cr 2n - j \cr} \right) \sum\limits_{x = 0}^b {\left( \matrix{ x \cr j \cr} \right)} } } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{ n \cr k \cr} \right\rangle } \right\rangle \left( \matrix{ k \cr 2n - j \cr} \right)\left( \matrix{ b + 1 \cr j + 1 \cr} \right)} } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{ n \cr k \cr} \right\rangle } \right\rangle \left( \matrix{ k + b + 1 \cr 2n + 1 \cr} \right)} \cr} $$