Use the ratio test to determine if $$ \sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n} n $$ ! converges or diverges and justify your answer. Answer Attempt 1 out of 2 Using the ratio test, $$ \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\square . \sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n} n $$ ! since the ratio is

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We are given the series

$$
\sum_{n=1}^{\infty} \left(\frac{5}{4}\right)^{n} n!
$$

with $$ a_n = \left(\frac{5}{4}\right)^{n} n! $$.

**Step 1. Compute the ratio $$\left|\frac{a_{n+1}}{a_n}\right|$$:**

$$
a_{n+1} = \left(\frac{5}{4}\right)^{n+1} (n+1)!
$$

Then

$$
\frac{a_{n+1}}{a_n} = \frac{\left(\frac{5}{4}\right)^{n+1}(n+1)!}{\left(\frac{5}{4}\right)^n n!} = \left(\frac{5}{4}\right) \cdot \frac{(n+1)!}{n!}
$$

Since $$\frac{(n+1)!}{n!} = n+1$$, we have

$$
\frac{a_{n+1}}{a_n} = \left(\frac{5}{4}\right)(n+1)
$$

Thus,

$$
\left|\frac{a_{n+1}}{a_n}\right| = \left(\frac{5}{4}\right)(n+1)
$$

**Step 2. Evaluate the limit:**

$$
L = \lim_{n\to\infty} \left(\frac{5}{4}\right)(n+1)
$$

Since

$$
\lim_{n\to\infty} (n+1) = \infty,
$$

we have

$$
L=\infty.
$$

**Step 3. Apply the Ratio Test:**

The ratio test states that if $$ L > 1 $$ (in fact, if $$ L = \infty $$), the series diverges.

Since $$ L = \infty > 1 $$, the series diverges.

**Conclusion:**

The series

$$
\sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n} n!
$$

diverges by the ratio test.
The series $$ \sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n} n! $$ diverges because the limit of the ratio $$ \left|\frac{a_{n+1}}{a_n}\right| $$ as $$ n $$ approaches infinity is infinity, which is greater than 1.

Use the ratio test to determine if ! converges or diverges
and justify your answer.
Answer Attempt 1 out of 2
Using the ratio test, !
since the ratio is

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Use the ratio test to determine if ! converges or diverges and justify your answer. Answer Attempt 1 out of 2 Using the ratio test, ! since the ratio is