Math Problem Statement

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

Solution

To determine whether the infinite series

n=13n2nnn!\sum_{n=1}^{\infty} \frac{3^{n^2}}{n^n \cdot n!}

converges, we'll apply the ratio test.

Ratio Test

For a series an\sum a_n, the ratio test says:

  • If limnan+1an=L\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L, then:
    • If L<1L < 1, the series converges.
    • If L>1L > 1, the series diverges.
    • If L=1L = 1, the test is inconclusive.

Here, an=3n2nnn!a_n = \frac{3^{n^2}}{n^n \cdot n!}.

We need to compute:

limnan+1an\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

Step 1: Write an+1a_{n+1}

We have:

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

Step 2: Form the ratio

Now, form the ratio an+1an\frac{a_{n+1}}{a_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.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Exponential Growth
Factorials
Ratio Test

Formulas

Ratio test: lim(n→∞) |a_{n+1}/a_n| = L
General form of the series: Σ (3^{n^2} / (n^n * n!))

Theorems

Ratio Test

Suitable Grade Level

Undergraduate Mathematics