sigma (infinity and n=1) 3^n^2/(n^n*n!) ratio test

To determine whether the infinite series

\[
\sum_{n=1}^{\infty} \frac{3^{n^2}}{n^n \cdot n!}
\]

converges, we'll apply the **ratio test**.

### Ratio Test

For a series \(\sum a_n\), the ratio test says:
- If \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L\), then:
  - If \(L < 1\), the series converges.
  - If \(L > 1\), the series diverges.
  - If \(L = 1\), the test is inconclusive.

Here, \(a_n = \frac{3^{n^2}}{n^n \cdot n!}\).

We need to compute:

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]

### Step 1: Write \(a_{n+1}\)

We have:

\[
a_{n+1} = \frac{3^{(n+1)^2}}{(n+1)^{n+1} \cdot (n+1)!}
\]

### Step 2: Form the ratio

Now, form the ratio \( \frac{a_{n+1}}{a_n} \):

\[
\frac{a_{n+1}}{a_n} = \frac{\frac{3^{(n+1)^2}}{(n+1)^{n+1} \cdot (n+1)!}}{\frac{3^{n^2}}{n^n \cdot n!}} 
= \frac{3^{(n+1)^2} \cdot n^n \cdot n!}{3^{n^2} \cdot (n+1)^{n+1} \cdot (n+1)!}
\]

Simplifying:

\[
= \frac{3^{n^2 + 2n + 1} \cdot n^n \cdot n!}{3^{n^2} \cdot (n+1)^{n+1} \cdot (n+1)!}
= 3^{2n+1} \cdot \frac{n^n \cdot n!}{(n+1)^{n+1} \cdot (n+1)!}
\]

### Step 3: Simplify the factorials and powers

Using the fact that \((n+1)! = (n+1) \cdot n!\), we get:

\[
= 3^{2n+1} \cdot \frac{n^n}{(n+1)^{n+1} \cdot (n+1)}
= 3^{2n+1} \cdot \frac{n^n}{(n+1)^{n+2}}
\]

### Step 4: Simplify the limit

Now take the limit as \(n \to \infty\):

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
= \lim_{n \to \infty} 3^{2n+1} \cdot \frac{n^n}{(n+1)^{n+2}}
\]

We can approximate \(\left(\frac{n}{n+1}\right)^n\) as \(e^{-1}\) as \(n \to \infty\). So:

\[
\frac{n^n}{(n+1)^{n+2}} \sim \frac{e^{-n}}{(n+1)^2}
\]

Thus, the limit becomes:

\[
\lim_{n \to \infty} 3^{2n+1} \cdot e^{-n} = \lim_{n \to \infty} \left( \frac{3^{2n+1}}{e^n} \right)
\]

Since \(3^{2n+1}\) grows much faster than \(e^n\), this limit tends to infinity:

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \infty
\]

### Conclusion:

Since the limit is greater than 1, the series **diverges** by the ratio test.

Would you like more details or have any questions?

### Related Questions:
1. What is the root test, and how does it compare with the ratio test?
2. How do factorials affect the convergence of series?
3. Can you apply the ratio test to a series involving logarithmic terms?
4. What happens when the ratio test gives a limit of 1?
5. How do growth rates of exponential functions influence series convergence?

### Tip:
In convergence tests, always simplify the ratio of consecutive terms before applying limits to avoid errors.

This page explains how to apply the ratio test to determine the convergence of the series Σ (3^n^2 / (n^n * n!)). Learn the step-by-step process, including simplifying powers and factorials, and see why the series diverges. This content is suitable for undergraduate-level mathematics students.

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