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.

  • 1
    $\begingroup$ For any $$\;n\in\Bbb N\;,\;\;\frac1{3n+2\cos n}\ge\frac1{3n+2}$$ and the series is, of course, a positive one. $\endgroup$
    – DonAntonio
    Aug 12, 2016 at 22:11

3 Answers 3


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.

  • $\begingroup$ Thank you for such a fast response. I realized I wasn't clear when I stated my question, I understand how we end up with 3/3 when we take the lim of the num and denom. My trouble is why does 2cos(n)/n equal to zero? After that it's just simple algebra 3/(3+0) = 1. $\endgroup$ Aug 14, 2016 at 0:04
  • $\begingroup$ It is not equal to zero. It just becomes ever smaller and converges to zero when $n$ grows larger and larger. After all, since $|\cos(n)| \leq 1$, we get $|2\cos(n)/n| \leq 2/n$, and since $2/n$ converges to $0$, $2\cos(n)/n$ does too. $\endgroup$
    – Anon
    Aug 14, 2016 at 0:22
  • $\begingroup$ Thank you so much. I had a feeling this was it. I asked my professor if we can take the absolute of value of cosine but he wasn't sure. I had a feeling this was it, thank you for clearing this up. $\endgroup$ Aug 14, 2016 at 3:39

$$\frac1{3n+2\cos n}\sim_\infty\frac1{3n},\quad\text{which diverge.}$$


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:

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


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