# Is $\sum_{n=1}^{\infty} \frac{ \sqrt{n + 1} (n + 1)! }{ e^{n + 5} }$convergent or divergent

I have worked on this my answer is L= Div and B

Consider the series $\displaystyle \sum_{n=1}^{\infty} a_n$ where

$$a_n = \frac{ \sqrt{n + 1} (n + 1)! }{ e^{n + 5} }$$

In this problem you must attempt to use the Ratio Test to decide whether the series converges.

Compute

$$L=\lim_{n\rightarrow\infty} \left| \frac{ a_{n+1} }{a_n} \right|$$ Enter the numerical value of the limit L if it converges, INF if it diverges to infinity, MINF if it diverges to negative infinity, or DIV if it diverges but not to infinity or negative infinity. L =

Which of the following statements is true?
A. The Ratio Test says that the series converges absolutely.
B. The Ratio Test says that the series diverges.
C. The Ratio Test says that the series converges conditionally.
D. The Ratio Test is inconclusive, but the series converges absolutely by another test or tests.
E. The Ratio Test is inconclusive, but the series diverges by another test or tests.
F. The Ratio Test is inconclusive, but the series converges conditionally by another test or tests.

My Work

$$\frac{ \sqrt{n + 2} (n + 2)! }{ e^{n + 6} }\frac{ e^{n + 5} }{ \sqrt{n + 1} (n + 1)! }$$ $$\frac{ \sqrt{n + 2} (n + 2) }{ e \sqrt{n+1} }$$ now limit and absolute value

$$\lim_{n\rightarrow\infty} \left| \frac{ \sqrt{n + 2} (n + 2) }{ e \sqrt{n+1} } \right|= \infty$$ Can you tell me why I am wrong.

Since $\displaystyle\lim_{n\to\infty}a_n = \infty$, your correct answer seems to be INF not DIV.