Calculus 2 infinite series $\frac{1}{3n + 2\cos n}$ The series is $\frac{1}{3n + 2\cos n}$ for $n$ from 1 to infinity.

Does the series converge or diverge? State the test applied.

I know $2\cos n$ will oscillate from -2 to 2. Divergence test give me 0 so it's inconclusive. I was suggested to use the limit comparison test. I was told to compare using $\frac{1}{3n}$. After computation the limit as $n$ goes to infinity looks like $\frac{3n}{3n + 2\cos n}$ which is equal to 1, this is what I don't understand: why does it equal 1?
Finally $\frac{1}{3n}$ is a harmonic series (factor the 1/3 out and you are left with $\frac{1}{n}$) which diverges. There fore we conclude the series diverges by limit comparison test.
 A: The sequence $\frac{3n}{3n+2\cos(n)}$ can be rewritten as $\frac{3}{3+2\cos(n)/n}$. Here you can easily see that the numerator converges to $3$ and the denominator also converges to $3$ (since the other summand converges to $0$), so the quotient converges to $1$.
The rest of your proof is correct.
A: For $n \geq 1$:
$$0<\frac{1}{3n+2} \leq \frac{1}{3n+2\cos n} \leq \frac{1}{3n-2}$$
Sum both sides of the inequality from $n=1$ to $\infty$ and notice that your bounds diverge too infinity as. So by the comparison test, the sum diverges.
Note we have:
$$\sum_{n=1}^{\infty} \frac{1}{3n-2}>\sum_{n=1}^{\infty} \frac{1}{3n} \to \infty$$
By term comparison. Also we have:
$$\sum_{n=1}^{\infty} \frac{1}{3n+2}>\sum_{n=1}^{\infty} \frac{1}{3n+3}=\frac{1}{3}\sum_{n=1}^{\infty} \frac{1}{n+1} \to \infty$$
We have the last sum above diverges to infinity by the comparison test with the harmonic series because:
$$\sum_{n=1}^{\infty} \frac{1}{n+1}>\sum_{n=2}^{\infty} \frac{1}{2n}$$
A: $$\frac1{3n+2\cos n}\sim_\infty\frac1{3n},\quad\text{which diverge.}$$
