Why does the series $\sum\limits_{n=2}^\infty\frac{\cos(n\pi/3)}{n}$ converge? Why does this series
$$\sum\limits_{n=2}^\infty\frac{\cos(n\pi/3)}{n}$$
converge? Can't you use a limit comparison with $1/n$? 
 A: First of all your conclusion is wrong since $\lim_{n \to \infty} \cos(n \pi/3)$ doesn't exist.
The convergence of $$\sum_{n=1}^N \dfrac{\cos(n\pi/3)}{n}$$ can be concluded based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.
Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.
First note that from Abel summation, we have that
\begin{align*}\sum_{n=1}^N a(n) b(n) &= \sum_{n=1}^N b(n)(A(n)-A(n-1))\\&= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\
&= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1} b(n+1)A(n) \\&= b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))\end{align*}
Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that
$$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$
Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n)  (b(n)-b(n+1))$$ converges absolutely. Now since
$$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges.
In your case, $a(n) = \cos(n \pi/3)$. Hence, $$A(N) = \displaystyle \sum_{n=1}^N a(n) = - \dfrac12 - \cos\left(\dfrac{\pi}3(N+2)\right)$$which is clearly bounded.
Also, $b(n) = \dfrac1{n}$ is a monotone decreasing sequence converging to $0$.
Hence, we have that $$\sum_{n=1}^N \dfrac{\cos(n\pi/3)}{n}$$ converges.

Look at some of my earlier answers for similar questions.
For what real numbers $a$ does the series $\sum \frac{\sin(ka)}{\log(k)}$ converge or diverge?
Give a demonstration that $\sum\limits_{n=1}^\infty\frac{\sin(n)}{n}$ converges.
If the partial sums of a $a_n$ are bounded, then $\sum{}_{n=1}^\infty a_n e^{-nt}$ converges for all $t > 0$

If you are interested in evaluating the series, here is a way out. We have for $\vert z \vert \leq 1$ and $z \neq 1$, $$\sum_{n=1}^{\infty} \dfrac{z^n}n = - \log(1-z)$$ Setting $z = e^{i \pi/3}$, we get that
$$\sum_{n=1}^{\infty} \dfrac{e^{in \pi/3}}n = - \log(1-e^{i \pi/3})$$
Hence,
\begin{align}
\sum_{n=1}^{\infty} \dfrac{\cos(n \pi/3)}n & = \text{Real part of}\left(\sum_{n=1}^{\infty} \dfrac{e^{in \pi/3}}n  \right)\\
& = \text{Real part of} \left(- \log(1-e^{i \pi/3}) \right)\\
& = - \log(\vert 1-e^{i \pi/3} \vert) = 0
\end{align}
Hence,
$$\sum_{n=2}^{\infty} \dfrac{\cos(n \pi/3)}n = - \dfrac{\cos(\pi/3)}1 = - \dfrac12$$
A: Note that $$\cos(n\pi/3) = 1/2, \  -1/2, \  -1, \  -1/2, \  1/2, \  1, \  1/2, \  -1/2, \  -1, \  \cdots $$ so your series is just 3 alternating (and convergent) series inter-weaved. Exercise: Prove that if $\sum a_n, \sum b_n$ are both convergent, then the sequence $$a_1, a_1+b_1, a_1+b_1+a_2, a_1+b_1+a_2+b_2, \cdots $$ is convergent. Applying that twice proves your series converges. 
