Determine convergence of $\frac{2n^3 + 7}{n^4 \sin^2 n}$

Question

Discuss the convergence of the series: $$\sum_{n=1}^\infty \frac{2n^3+7}{n^4 \sin^2 n}$$

My approach: Let, $b_n=\frac{2n^3+7}{n^4\sin^2 n}$ $$\sin^2n=\frac{1-\cos 2n}{2}$$ Let,$a_n=\frac{1}{n^4}$

Using limit comparison test, $$\lim_{n \to \infty}\frac{a_n}{b_n}=\lim_{n \to \infty}\frac{\frac{1}{n^4}}{\frac{2n^3+7}{n^4\sin^2 n}}=\lim_{n \to \infty}\frac{\sin^2n}{2n^3+7}=\lim_{n \to \infty}\frac{1-\cos2n}{2n^3+7}=\lim_{n \to \infty}\frac{1}{2n^3+7}-\frac{\cos2n}{2n^3+7}$$ $$\lim_{n \to \infty}\frac{1}{2n^3+7}-\frac{\cos2n}{2n^3+7}=0-\lim_{n \to \infty}\frac{\cos 2n}{2n^3+7}$$

This next step is what I'm very unsure of. Although the limit of x as $\cos x$ tends to infinity isn't defined, it won't exceed $+1$, and wont drop below $-1$. So, $2n^3+7$ becomes the deciding factor again, and that whole limit turns out to be zero.

Therefore, $$\lim_{n \to \infty}\frac{a_n}{b_n}=0$$

According to limit comparison test, if the limit equals zero, and $a_n$ converges, then $b_n$ also converges. WKT $\frac{1}{n^4}$ is convergent, therefore $b_n$ is also convergent.

So, this was my approach, and I'm not very sure of my method, as well as answer. If anyone has a better way to solve this question(which I'm damn sure, exists), kindly let me know. Thanks, in advance!!

Answer 1

We have

$$\frac{2n^3+7}{n^4 \sin^2 n}\geq \frac{2n^3+0}{n^4 \cdot 1}=\frac{2}{n}$$

Thus, the sum

$$\sum_{n=1}^\infty \frac{2n^3+7}{n^4 \sin^2 n}\geq \sum_{n=1}^\infty \frac{2}{n}=\infty$$

and diverges.

Answer 2

This series is in fact divergent. In general it is not true that $\lim_{n\to\infty}\left(\frac{a_n}{b_n}\right)=0$ and $\sum_na_n$ converges implies $\sum_nb_n$ converges. xbh provides an intuitive counter example

Note that for any $n\in\mathbb{N}$, $b_n\geq\frac{2}{n}$, and so

$$\sum_{n=1}^{\infty}\frac{2n^3+7}{n^4\sin^2(n)}\geq\sum_{n=1}^{\infty}\frac{2}{n}$$.

But $$\sum_{n=1}^{\infty}\frac{2}{n}$$

is just twice the harmonic series, which diverges. Hence, by the comparison test, your original series diverges.