Study the convergence of this: $\sum_n \frac{n! }{ 6\cdot7\cdots(n+5)}$ [closed]
Firstly, we have $$ \sum_{i=1}^n\frac{i!}{6\cdot 7 \cdot \cdots \cdot (i+5)} = \sum_{i=1}^n \frac{5!}{(i+1)(i+2)(i+3)(i+4)(i+5)} \\ \le \sum_{i=1}^n \frac{5!}{(i+1)(i+2)} \le \sum_{i=1}^n \Big[\frac{5!}{(i+1)} -\frac{5!}{(i+2)}\Big] = \frac{5!}{2} - \frac{5!}{n+2} \to \frac{5!}{2} $$ Since the sum is upperbounded, it's convergent.