$\text { Determine if the series } \sum _ { n = 1 } ^ { \infty } a _ { n } \text { defined by the formula converges or diverges. }$
-$$a _ { 1 } = 4 , a _ { n + 1 } = \frac { 4 n - 6 \sin n } { 2 n - 3 \cos n } a _ { n }$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

We can use the comparison test to prove that the series diverges. Note that for all $n \geq 2$,
\begin{align*}
a_{n+1} &= \frac{4n-6\sin(n)}{2n-3\cos(n)}a_n \
&= \frac{(4n-6\sin(n))(2n-3\cos(n-1))}{(2n-3\cos(n))(2n-3\cos(n-1))}\cdot\frac{(2n-3\cos(n))a_n}{a_n} \
&= \frac{(8n^2-24n\sin(n)-12n\cos(n)+18\sin(n)\cos(n)-18\sin(n)+9\cos^2(n-1))}{(4n^2-12n\cos(n)+9\cos^2(n))(2n-3\cos(n-1))}a_n.
\end{align*}
We will now show that $\frac{a_{n+1}}{a_n} \geq \frac{1}{2}$ for all $n$ sufficiently large. Observe that
$$\frac{8n^2-24n\sin(n)-12n\cos(n)+18\sin(n)\cos(n)-18\sin(n)+9\cos^2(n-1)}{4n^2-8n\cos(n)} \geq \frac{16n^2-24n}{4n^2} \geq \frac{1}{2}$$
for every $n$ such that $\cos(n) \leq -1/2$. There are infinitely many such $n$ by the periodicity of the cosine function, so we can choose $n_0$ large enough such that $\cos(n) \leq -1/2$ for all $n \geq n_0$. Thus, for all $n \geq n_0$,
$$\frac{a_{n+1}}{a_n} \geq \frac{1}{2}.$$
Since $\sum_{n=1}^\infty 4$ diverges, it follows that $\sum_{n=1}^\infty a_n$ also diverges by the ratio test. Therefore, the answer is $\boxed{\textbf{(A)}}$.

Determine if the series ∑n=1∞an defined by the formula converges or diverges. \text { Determine if the series } \sum _ { n = 1 } ^ { \infty } a _ { n } \text { defined by the formula converges or diverges. } Determine if the series ∑n=1∞​an​ defined by the formula converges or diverges.

$\text { Determine if the series } \sum _ { n = 1 } ^ { \infty } a _ { n } \text { defined by the formula converges or diverges. }$