Test the convergence of the series 3/4 + 3.4/4.6 + 3.4.5/4.6.8 + ...
Understand the Problem
The question is asking to test the convergence of a specific series. It provides the first few terms of the series, which involves fractions with numerators and denominators that appear to follow a pattern. The goal is to determine whether the series converges or diverges.
Answer
The series converges.
Answer for screen readers
The series converges.
Steps to Solve
- Identify the general term of the series
The terms of the series can be expressed as:
$$ a_n = \frac{3 \cdot (n + 2)}{4 \cdot (n + 3)(n + 1)} $$
Where ( n ) starts from 1.
- Simplify the general term
Next, we simplify ( a_n ):
$$ a_n = \frac{3(n + 2)}{4(n + 3)(n + 1)} $$
- Use the Comparison Test for convergence
To test for convergence, we compare this series with a known convergent series. We can compare it to ( \frac{1}{n^2} ) since it converges and behaves similarly for large ( n ).
- Evaluate the limit for the limit comparison test
We need to evaluate the limit:
$$ L = \lim_{n \to \infty} \frac{a_n}{\frac{1}{n^2}} = \lim_{n \to \infty} 3n^2 \cdot \frac{(n+2)}{4(n+3)(n+1)} $$
- Calculate the limit
As ( n \to \infty ),
$$ L = \lim_{n \to \infty} \frac{3n^2(n + 2)}{4(n^2 + 4n + 3)} = \lim_{n \to \infty} \frac{3(n + 2)}{4(n + 4/n + 3/n^2)} = \frac{3}{4} $$
- Determine convergence based on the limit
Since ( L = \frac{3}{4} ) is a positive finite number, by the Limit Comparison Test, the original series converges if the comparison series converges.
The series converges.
More Information
The Limit Comparison Test is a powerful method for determining the convergence of series, particularly effective when one series mimics the behavior of another well-known series.
Tips
- Not simplifying the general term accurately, leading to incorrect comparisons.
- Confusing the behavior of series with different rates of growth.
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