Software for evaluating summation expressions involving Stirling numbers of the first kind. 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!
 A: 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 $x$
$$
\eqalign{
  & \sum\limits_{x = 0}^b {\left[ \matrix{  x \cr   x - n \cr}  \right]}
  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
 {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {\left\langle {\left\langle \matrix{  n \cr   k \cr}  \right\rangle } \right\rangle
 \left( \matrix{  k \cr   2n - j \cr}  \right)
 \sum\limits_{x = 0}^b {\left( \matrix{  x \cr  j \cr}  \right)} } }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
 {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {\left\langle {\left\langle \matrix{  n \cr   k \cr}  \right\rangle } \right\rangle
 \left( \matrix{  k \cr   2n - j \cr}  \right)\left( \matrix{  b + 1 \cr   j + 1 \cr}  \right)} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
 {\left\langle {\left\langle \matrix{  n \cr   k \cr}  \right\rangle } \right\rangle
 \left( \matrix{  k + b + 1 \cr   2n + 1 \cr}  \right)}  \cr} 
$$
