To analyze the sequence $a_n = \frac{(-1)^{n+1} \cdot 2n}{n^2 + 3n + 4}$, we first calculate the first six terms by substituting $n = 1, 2, 3, \dots, 6$. Then, we'll evaluate if the sequence has a limit.

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Step 1: Calculate the First Six Terms

For $n = 1$:

$a_1 = \frac{(-1)^{1+1} \cdot 2 \cdot 1}{1^2 + 3 \cdot 1 + 4} = \frac{1 \cdot 2}{1 + 3 + 4} = \frac{2}{8} = 0.25$

For $n = 2$:

$a_2 = \frac{(-1)^{2+1} \cdot 2 \cdot 2}{2^2 + 3 \cdot 2 + 4} = \frac{-1 \cdot 4}{4 + 6 + 4} = \frac{-4}{14} = -\frac{2}{7}$

For $n = 3$:

$a_3 = \frac{(-1)^{3+1} \cdot 2 \cdot 3}{3^2 + 3 \cdot 3 + 4} = \frac{1 \cdot 6}{9 + 9 + 4} = \frac{6}{22} = \frac{3}{11}$

For $n = 4$:

$a_4 = \frac{(-1)^{4+1} \cdot 2 \cdot 4}{4^2 + 3 \cdot 4 + 4} = \frac{-1 \cdot 8}{16 + 12 + 4} = \frac{-8}{32} = -\frac{1}{4}$

For $n = 5$:

$a_5 = \frac{(-1)^{5+1} \cdot 2 \cdot 5}{5^2 + 3 \cdot 5 + 4} = \frac{1 \cdot 10}{25 + 15 + 4} = \frac{10}{44} = \frac{5}{22}$

For $n = 6$:

$a_6 = \frac{(-1)^{6+1} \cdot 2 \cdot 6}{6^2 + 3 \cdot 6 + 4} = \frac{-1 \cdot 12}{36 + 18 + 4} = \frac{-12}{58} = -\frac{6}{29}$

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First Six Terms:

$a_1 = 0.25, \quad a_2 = -\frac{2}{7}, \quad a_3 = \frac{3}{11}, \quad a_4 = -\frac{1}{4}, \quad a_5 = \frac{5}{22}, \quad a_6 = -\frac{6}{29}$

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Step 2: Determine If the Sequence Has a Limit

To evaluate if the sequence has a limit, examine the behavior of $a_n$ as $n \to \infty$.

The general term is: $a_n = \frac{(-1)^{n+1} \cdot 2n}{n^2 + 3n + 4}.$

• As $n \to \infty$, the denominator $n^2 + 3n + 4$ grows much faster than the numerator $2n$. Thus, the magnitude of $a_n$ decreases to $0$.

Additionally, since the numerator alternates between positive and negative due to the factor $(-1)^{n+1}$, the sequence oscillates around $0$, but its oscillations diminish in size.

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Conclusion:

The sequence does not converge to a single value (no limit in the conventional sense) but instead approaches $0$ in magnitude while oscillating. We say the sequence does not have a limit because of this oscillatory behavior.

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Do you want further clarification on limits or sequences?
Here are five related questions to explore:

1. What is the difference between oscillatory and convergent sequences?
2. Can sequences with alternating signs still have limits?
3. How do you determine if a sequence converges absolutely?
4. Can you find a sequence with a similar form but that converges?
5. What happens if you remove the oscillation (e.g., replace $(-1)^{n+1}$ with 1)?

Tip: To determine a sequence's limit, focus on the dominant terms in the numerator and denominator as $n \to \infty$.