\( a_{n}=\frac{(-1)^{n}}{(n+1)(n+3)} \) 1) Find \( \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \) \( \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\square \)

Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(-1\right)^{n+1}}{\left(n+2\right)\left(n+4\right)}\times \frac{\left(n+1\right)\left(n+3\right)}{\left(-1\right)^{n}}\right|\right)\) - step1: Multiply the terms: \(\lim _{n\rightarrow +\infty}\left(\left|-\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\) - step2: Calculate the absolute value: \(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\) - step3: Rewrite the expression: \(|\lim _{n\rightarrow +\infty}\left(\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right)|\) - step4: Calculate: \(|1|\) - step5: Calculate the absolute value: \(1\) Calculate the limit \( \lim_{n \rightarrow \infty}\left|\frac{(-1)^{n+1}}{(n+2)(n+4)} \cdot \frac{(n+1)(n+3)}{(-1)^{n}}\right| \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(-1\right)^{n+1}}{\left(n+2\right)\left(n+4\right)}\times \frac{\left(n+1\right)\left(n+3\right)}{\left(-1\right)^{n}}\right|\right)\) - step1: Multiply the terms: \(\lim _{n\rightarrow +\infty}\left(\left|-\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\) - step2: Calculate the absolute value: \(\lim _{n\rightarrow +\infty}\left(\left|\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right|\right)\) - step3: Rewrite the expression: \(|\lim _{n\rightarrow +\infty}\left(\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+4\right)}\right)|\) - step4: Calculate: \(|1|\) - step5: Calculate the absolute value: \(1\) To find \( \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \), we start by substituting the expression for \( a_n \): \[ a_n = \frac{(-1)^{n}}{(n+1)(n+3)} \] Now, we calculate \( a_{n+1} \): \[ a_{n+1} = \frac{(-1)^{n+1}}{(n+2)(n+4)} \] Next, we find the ratio \( \frac{a_{n+1}}{a_n} \): \[ \frac{a_{n+1}}{a_n} = \frac{\frac{(-1)^{n+1}}{(n+2)(n+4)}}{\frac{(-1)^{n}}{(n+1)(n+3)}} = \frac{(-1)^{n+1}}{(-1)^{n}} \cdot \frac{(n+1)(n+3)}{(n+2)(n+4)} = -\frac{(n+1)(n+3)}{(n+2)(n+4)} \] Taking the absolute value: \[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)(n+3)}{(n+2)(n+4)} \] Now, we compute the limit as \( n \) approaches infinity: \[ \lim_{n \rightarrow \infty} \frac{(n+1)(n+3)}{(n+2)(n+4)} \] Dividing the numerator and the denominator by \( n^2 \): \[ = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{\frac{2}{n} + \frac{4}{n^2}} = \frac{1 \cdot 1}{1 \cdot 1} = 1 \] Thus, we have: \[ \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| = 1 \] So the final answer is: \[ \lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| = 1 \]