1. Put $B(n,k)=k! S(n,k)$, here $S(n,k)$ be the Srirling numbers of second kind.


Prove that 
$$
\sum\_{j=i+1}^n \frac{(-1)^{j+1-i}}{j-i} B(n,j)= n B(n-1,i),i=0,1,\ldots,n-1
$$


1. Put $T(n,i)=\displaystyle \sum\_{j=i}^n (-1)^{j-i} 2^{n-j} B(n,j) {j-1 \choose i-1}$.


Prove that 


$$
\sum\_{j=i+1}^n \frac{1-(-1)^{j-i}}{2(j-i)}T(n,j)= n T(n-1,i),i=0 \ldots n-1.
$$



How to prove  recursion formulas  for Stirling  numbers?
Here is a proof of the first formula. We use the exponential generating function
$\sum\_{n=0}^\infty S(n,k) x^n/n! = (e^x-1)^k/k!$.


Multiply the left side by $x^n/n!$ and sum on $n$ from 1 to $\infty$. We obtain
\begin{align\*}
\sum\_{j=i+1}^\infty\frac{(-1)^{j+1-i}}{j-i} (e^x-1)^j
 &=\sum\_{k=1}^\infty \frac{(-1)^{k-1}}{k}(e^x-1)^{i+k}\\
 &=(e^x-1)^i \sum\_{k=1}^\infty \frac{(-1)^{k-1}}{k}(e^x-1)^{k}\\
 &=(e^x-1)^i\log(1+(e^x-1))= x(e^x-1)^i\\
 &=\sum\_{n=1}^\infty n\cdot i!\, S(n-1,i)\frac{x^n}{n!}.
\end{align\*}


It seems likely that a similar approach will work for the second formula, but I have not tried it.


