Question

1. -/1.11 pointsdetermine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.){n²e⁻³ⁿ}limₙ→∞ aₙ =14. -/1.11 pointsdetermine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.)aₙ = cos²(n)/2ⁿlimₙ→∞ aₙ =

Explanation:

Step1: Analyze the sequence \(\{n^{2}e^{-3n}\}\)

We know that \(e^{-3n}=\frac{1}{e^{3n}}\). So, we need to find \(\lim_{n
ightarrow\infty}\frac{n^{2}}{e^{3n}}\).
Since we have the indeterminate form \(\frac{\infty}{\infty}\) (as \(n
ightarrow\infty\), \(n^{2}
ightarrow\infty\) and \(e^{3n}
ightarrow\infty\)), we can apply L - H rule.
Let \(f(n) = n^{2}\), \(g(n)=e^{3n}\). Then \(f^{\prime}(n) = 2n\), \(g^{\prime}(n)=3e^{3n}\).
So, \(\lim_{n
ightarrow\infty}\frac{n^{2}}{e^{3n}}=\lim_{n
ightarrow\infty}\frac{2n}{3e^{3n}}\) (still \(\frac{\infty}{\infty}\) form).

Step2: Apply L - H rule again

Now, for \(\lim_{n
ightarrow\infty}\frac{2n}{3e^{3n}}\), let \(u(n)=2n\), \(v(n) = 3e^{3n}\). Then \(u^{\prime}(n)=2\), \(v^{\prime}(n)=9e^{3n}\).
So, \(\lim_{n
ightarrow\infty}\frac{2n}{3e^{3n}}=\lim_{n
ightarrow\infty}\frac{2}{9e^{3n}}\)

Step3: Evaluate the limit

As \(n
ightarrow\infty\), \(e^{3n}
ightarrow\infty\). So, \(\lim_{n
ightarrow\infty}\frac{2}{9e^{3n}} = 0\)

for the second sequence \(\{a_{n}=\frac{\cos^{2}(n)}{2^{n}}\}\)

Step1: Use the Squeeze Theorem

We know that \(0\leqslant\cos^{2}(n)\leqslant1\) for all \(n\). Then \(\frac{0}{2^{n}}\leqslant\frac{\cos^{2}(n)}{2^{n}}\leqslant\frac{1}{2^{n}}\)

Step2: Evaluate the limits of the bounding sequences

\(\lim_{n
ightarrow\infty}\frac{0}{2^{n}}=0\) and \(\lim_{n
ightarrow\infty}\frac{1}{2^{n}}=\lim_{n
ightarrow\infty}(\frac{1}{2})^{n}=0\) (since \(|r|=\frac{1}{2}<1\) and \(\lim_{n
ightarrow\infty}r^{n}=0\) for \(|r|<1\))

Answer:

For the sequence \(\{n^{2}e^{-3n}\}\), \(\lim_{n
ightarrow\infty}a_{n}=0\)
For the sequence \(\{a_{n}=\frac{\cos^{2}(n)}{2^{n}}\}\), \(\lim_{n
ightarrow\infty}a_{n}=0\)