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...
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...
Copyright © 2021 JogjaFile Inc.
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.