Let $i$, $k$ be integers such that $2 \leq i \leq k$. I would like to show that the sum $$ \sum_{j=1}^{i1} \frac{(1)^{j1}(ij)^k}{(ij)! (j1)!} $$ is positive. I have carried out extensive numerical experiments to check this for small values of $k$. In fact, much more should be true. Define polynomials $$ U(x)=(x+i1)^k $$ and $$ V(x)=x(x+1)\cdots(x+i1). $$ Let $Q$ and $R$ be the quotient and remainder on dividing $U$ by $V$. The above sum is the leading coefficient of $R$. It seems that all the coefficients of $Q$ and $R$ are always positive, and it would be nice to prove this, but I only need the positivity of the above sum. This question has applications for proving the irrationality of certain series.
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

These are Stirling numbers of the second kind. More precisely your sum is S(k,i1) where $S$ denotes Stirling number of the second kind. 


More general result is the following: if $R$ is remainder of $x^k$ modulo $(xc_1)(xc_2)\dots (xc_i)$ with nonnegative $c_i$'s, then leading coefficient of $R$ is positive. Indeed, let $R=ax^{i1}+\dots$, then $f(x):=x^kax^{i1}\dots$ has roots in $c_i$'s, then by Rolle theorem $f^{(i1)}$ has at least one positive root, which is true iff $a>0$. 

