Proving divergence of an alternating series

Question

Consider following series $$\sum_{n=0}^\infty \frac{(3-(-1)^n)\cos((n-1)\pi)}{2n}.$$ According to answers in my textbook, this is a divergent series. The problem is, I don't know how to prove it. I asked the teacher, who said to write it like this $$\sum_{n=1}^\infty \frac{(3-(-1)^n)(-1)^{n+1}}{2n}=\sum_{n=1}^\infty(-1)^{n+1}\frac{3}{2n}+\frac{1}{2n},$$ and then to split the series into a sum of two: $$\sum_{n=1}^\infty{(-1)^{n+1}\frac{3}{2n}} + \sum_{n=1}^\infty{\frac{1}{2n}}.$$ Now, the first one is convergent, and the second one is divergent, hence the series in question is also divergent. But I don't believe this is correct. Isn't this the case that $\sum{a_n+b_n} = \sum{a_n}+\sum{b_n} \iff \sum{a_n}$ converges and $\sum{b_n}$ converges? Is the argument presented valid and if not how can I prove the divergence of this series?

Answer 1

The proof is basically correct. The strictly correct lemma to use is sum rule:

(Sum rule) If $A=\sum a_n$ and $B=\sum b_n$ converge, then so does $C=\sum(a_n+b_n)$ and the answer is $C=A+B$.

So suppose the sum in question converged, let's call it $A=\sum a_n$. Then as $B=- \sum_{n=1}^\infty{(-1)^{n+1}\frac{3}{2n}} =\sum b_n$ converges, Sum rule gives that the sum of terms $a_n+b_n$ converges. But this is the above multiple of the harmonic series; we have reached a contradiction, so our assumption that $A$ converged was false.