To determine the convergence of the given series

$\sum_{n=1}^\infty \frac{(n+2)!}{(n+5)^n},$

let's analyze it in detail:

Step 1: Apply Ratio Test

The ratio test is a common way to check the convergence of series involving factorials and powers. The test computes:

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|,$

where $a_n = \frac{(n+2)!}{(n+5)^n}$.

Compute $\frac{a_{n+1}}{a_n}$:

$a_n = \frac{(n+2)!}{(n+5)^n}, \quad a_{n+1} = \frac{(n+3)!}{(n+6)^{n+1}}.$

The ratio is: $\frac{a_{n+1}}{a_n} = \frac{\frac{(n+3)!}{(n+6)^{n+1}}}{\frac{(n+2)!}{(n+5)^n}} = \frac{(n+3)!}{(n+2)!(n+6)^{n+1}} \cdot (n+5)^n.$

Simplify $(n+3)!$: $(n+3)! = (n+3)(n+2)!.$

Thus: $\frac{a_{n+1}}{a_n} = \frac{(n+3)(n+2)!}{(n+2)!(n+6)^{n+1}} \cdot (n+5)^n = \frac{n+3}{(n+6)^{n+1}} \cdot (n+5)^n.$

Split the powers of $(n+6)$ and $(n+5)$:

$\frac{a_{n+1}}{a_n} = \frac{n+3}{n+6} \cdot \left( \frac{n+5}{n+6} \right)^n.$

Step 2: Take the Limit as $n \to \infty$

As $n \to \infty$:

1. $\frac{n+3}{n+6} \to 1$,
2. $\left( \frac{n+5}{n+6} \right)^n \to e^{-1}$ because $\ln\left( 1 - \frac{1}{n+6} \right) \sim -\frac{1}{n+6}$.

Thus: $L = 1 \cdot e^{-1} = \frac{1}{e}.$

Step 3: Interpret the Result

From the ratio test:

• If $L < 1$, the series converges absolutely.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

Here, $L = \frac{1}{e} < 1$. Therefore, the series converges absolutely.

Final Conclusion

The series

$\sum_{n=1}^\infty \frac{(n+2)!}{(n+5)^n}$

converges absolutely.

Would you like more details about any of these steps or examples of other series tests?

Related Questions

1. What happens if $L = 1$ in the ratio test?
2. How does the root test compare to the ratio test for factorial series?
3. What are alternating series, and how is their convergence tested?
4. Can this series be approximated using dominant terms for $n \to \infty$?
5. How does absolute convergence differ from conditional convergence?

Tip

For series involving factorials, always consider the ratio test or Stirling's approximation for large $n$.