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Math  /  Calculus

Question∑n=1∞1⋅4⋅7⋅…⋅(3n−2)2⋅6⋅10⋅…⋅(4n−2)\sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{2 \cdot 6 \cdot 10 \cdot \ldots \cdot(4 n-2)}

Studdy Solution

STEP 1

1. The series is an infinite series starting from n=1 n = 1 .
2. The numerator of each term is a product of an arithmetic sequence with the first term as 1 and a common difference of 3.
3. The denominator of each term is a product of an arithmetic sequence with the first term as 2 and a common difference of 4.
4. We need to determine the convergence or divergence of the series.

STEP 2

1. Identify the general term of the series.
2. Simplify the general term.
3. Determine the convergence or divergence of the series using a suitable test.

STEP 3

Identify the general term of the series. The general term is given by:
an=1⋅4⋅7⋅…⋅(3n−2)2⋅6⋅10⋅…⋅(4n−2) a_n = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3n-2)}{2 \cdot 6 \cdot 10 \cdot \ldots \cdot(4n-2)}

STEP 4

Simplify the general term. First, recognize the pattern in the numerator and denominator:
- The numerator is a product of terms in the sequence 1,4,7,…,(3n−2) 1, 4, 7, \ldots, (3n-2) . - The denominator is a product of terms in the sequence 2,6,10,…,(4n−2) 2, 6, 10, \ldots, (4n-2) .
The number of terms in each product is n n .

STEP 5

Express the product in the numerator and denominator using factorials or Pochhammer symbols if possible.
For the numerator: ∏k=1n(3k−2) \prod_{k=1}^{n} (3k-2)
For the denominator: ∏k=1n(4k−2) \prod_{k=1}^{n} (4k-2)

STEP 6

Determine the convergence or divergence of the series. We can use the ratio test or another suitable test for convergence.
Apply the ratio test:
lim⁡n→∞∣an+1an∣ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
Calculate:
an+1=1⋅4⋅7⋅…⋅(3(n+1)−2)2⋅6⋅10⋅…⋅(4(n+1)−2) a_{n+1} = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3(n+1)-2)}{2 \cdot 6 \cdot 10 \cdot \ldots \cdot(4(n+1)-2)}
=1⋅4⋅7⋅…⋅(3n+1)2⋅6⋅10⋅…⋅(4n+2) = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3n+1)}{2 \cdot 6 \cdot 10 \cdot \ldots \cdot(4n+2)}
an+1an=(3n+1)(4n+2) \frac{a_{n+1}}{a_n} = \frac{(3n+1)}{(4n+2)}
lim⁡n→∞∣3n+14n+2∣=lim⁡n→∞3n+14n+2=34 \lim_{n \to \infty} \left| \frac{3n+1}{4n+2} \right| = \lim_{n \to \infty} \frac{3n+1}{4n+2} = \frac{3}{4}
Since 34<1 \frac{3}{4} < 1 , the series converges.
The series converges.

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