Study the convergence of this: $\sum_n \frac{n! }{ 6\cdot7\cdots(n+5)}$

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convergence for this one: $$\sum_n \frac{n! }{ 6\cdot7\cdots(n+5)}$$

I tried to calculate it standard and didn't got an answer...

Tried D'Alembert but didn't work.

Thank you very much for all the replies!!! I appreciate! It was very obvious...

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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.

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Your serie is equal to $$\sum_0^\infty {(1*2*3*4*5) \over (n+1)(n+2)(n+3)(n+4)(n+5)}<120\sum_0^\infty { 1 \over (n+1)^5}$$ which trivially converges.