On what b values does the series converge? On what $b$ value does the series converge? $$\sum_{k=1}^∞  \frac{k!}{(b+1)(b+2)...(b+k)}$$ I know that in order for the series to converge, the limit has to be equal to zero. By using D'Alembert's sequence I calculated, that $$\sum_{k=1}^∞  \frac{k!}{(b+1)(b+2)...(b+k)} = \lim_{k\to ∞} |\frac{1}{b+k}|$$.
Does that mean when $b \in \mathbb N$, the series converges?

I don't think the ratio test will help (if this is what you mean by "d'Alembert's sequence"). If we tried to use it, we would see $$\left \lvert \frac{a_{k+1}}{a_k} \right \rvert = \frac{(k+1)!}{(b+1)(b+2)\cdots(b+k)(b+k+1)}  \frac{(b+1)(b+2)\cdots(b+k)}{k!} = \frac{k+1}{b+k+1} \to 1$$ as $k\to \infty$ so the test is inconclusive. 
Certainly the series does not converge for $b = 1$ since then we have $$\frac{k!}{(2)(3)\cdots(k+1)} = \frac{1}{k+1}$$ so the series is the Harmonic series (with the first term removed). Similarly, if $b \in \mathbb N$, $b > 1$, then for large enough $k$,  $$\frac{k!}{(b+1)(b+2)\cdots(b+k)} = \frac{b!}{(k+1)\cdots(k+b)} = \frac{b!}{k^b + \text{ other positive terms}} =  \mathcal O(k^{-b})$$ so the series converges by comparison with $\sum \frac{1}{k^b}$. Since the coefficients $$ a_{k,b} =\frac{k!}{(b+1)(b+2)\cdots(b+k)}$$ are decreasing in $b$, this shows that we have convergence for any $b \ge 2$ and divergence for any $b \le 1$. I suspect that the estimate $$a_{k,b} = \mathcal O(k^{-b}) \,\,\,\, \text{ for } k \gg 1$$ also holds for non-integer $b$ and thus we have convergence for any $b > 1$ but the proof of this fact eludes me.
EDIT: Indeed, if we multiply the top and bottom by $b!$, we have $$\sum^\infty_{k=1} \frac{k!}{(b+1)(b+2)\cdots(b+k)} = \sum^\infty_{k=1} \frac{k! b!}{(k+b)!} = \sum^\infty_{k=1} \frac{1}{\binom{k+b}b}$$ and thus using $$\binom{n}m \ge \frac{n^m}{m^m},$$ which holds for all $0 \le m \le n$ (including non-integer values if we interpret factorials using the $\Gamma$ function), we see $$\sum^\infty_{k=1} \frac{k!}{(b+1)(b+2)\cdots(b+k)} \le b^b \sum^\infty_{k=1}\frac{1}{(k+b)^b}\le b^b \sum^\infty_{k=1}\frac{1}{k^b}.$$ Thus we do indeed have convergence whenever $b > 1$. 
