Math Problem Statement

Study the convergence of the following series: a) P∞ n=1 a n (n!)1/n

Solution

We are tasked with studying the convergence of the series:

n=1an(n!)1/n\sum_{n=1}^{\infty} \frac{a_n}{(n!)^{1/n}}

To analyze the convergence of this series, we'll employ some standard tests for series convergence. We assume that the terms ana_n satisfy certain conditions, but for general purposes, we'll first focus on the form involving n!n!.

Step 1: Rewrite the general term

The general term of the series is:

bn=an(n!)1/nb_n = \frac{a_n}{(n!)^{1/n}}

Step 2: Applying the ratio test

One of the most useful tests for series involving factorials is the ratio test, which states that a series bn\sum b_n converges if:

L=limnbn+1bnL = \lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right|

exists and satisfies L<1L < 1, and diverges if L>1L > 1. If L=1L = 1, the test is inconclusive.

Let’s compute this limit for our series.

= \frac{a_{n+1}}{a_n} \cdot \frac{(n!)^{1/n}}{((n+1)!)^{1/(n+1)}}$$ Now simplify the factorial term: $$\frac{(n!)^{1/n}}{((n+1)!)^{1/(n+1)}} = \frac{(n!)^{1/n}}{((n+1) \cdot n!)^{1/(n+1)}} = \frac{(n!)^{1/n}}{(n+1)^{1/(n+1)} \cdot (n!)^{1/(n+1)}}$$ $$= \frac{1}{(n+1)^{1/(n+1)}} \cdot \frac{(n!)^{1/n}}{(n!)^{1/(n+1)}}$$ The second term simplifies to: $$\frac{(n!)^{1/n}}{(n!)^{1/(n+1)}} = (n!)^{\frac{1}{n} - \frac{1}{n+1}}$$ Since $$\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}$$, we have: $$(n!)^{\frac{1}{n} - \frac{1}{n+1}} = (n!)^{\frac{1}{n(n+1)}}$$ For large $$n$$, this expression behaves like 1 (since $$\frac{1}{n(n+1)}$$ tends to 0 as $$n \to \infty$$). Thus, for large $$n$$, we are left with: $$\frac{b_{n+1}}{b_n} \sim \frac{a_{n+1}}{a_n} \cdot \frac{1}{(n+1)^{1/(n+1)}}$$ For large $$n$$, $$(n+1)^{1/(n+1)} \approx 1$$, so the limit simplifies to: $$\lim_{n \to \infty} \frac{b_{n+1}}{b_n} \sim \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$$ ### Step 3: Conclusion - If $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L < 1$$, then the series converges. - If $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L > 1$$, then the series diverges. - If $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L = 1$$, the test is inconclusive, and further analysis may be required (perhaps using other methods like the root test). ### Summary The convergence of the series depends on the behavior of the sequence $$a_n$$. Specifically: - If $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1$$, the series converges. - If $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} > 1$$, the series diverges. Would you like details on testing a specific form of $$a_n$$? ### 5 Related Questions: 1. How does the root test compare to the ratio test for factorial terms? 2. What are some typical examples of sequences $$a_n$$ that make the series converge? 3. Can we use Stirling's approximation to simplify the analysis of factorial terms? 4. How does the behavior of $$a_n$$ affect the radius of convergence in power series? 5. What is the role of the alternating series test in convergence? **Tip:** The ratio test is especially effective for series involving factorials or exponential terms due to the rapid growth of these functions.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Ratio Test
Factorials

Formulas

b_n = a_n / (n!)^{1/n}
Ratio Test: L = lim (n -> infinity) |b_{n+1} / b_n|

Theorems

Ratio Test
Root Test

Suitable Grade Level

Undergraduate - Advanced Calculus/Real Analysis