Prove that $\sum_{n=1}^{\infty}\frac{(3-(-1)^n)\cos(n-1)\pi}{2n}$ is divergent

Question

Prove that $\displaystyle\sum_{n=1}^{\infty}\frac{(3-(-1)^n)\cos(n-1)\pi}{2n}$ is divergent.

I tried:

1. Limit of the summand is equal 0, won't help.

2. I thought of Dirichlet's test (sequence of partial sums of $[3-(-1)^n]\cos(n-1)\pi$ is not bounded) but I believe it works only one way (for proving that series converges).

Answer 1

It diverges because if you add the convergent $\frac{(-1)^n}{n}$ to the general term, you get the series $(\frac{1}{2n-1})_{n=1}^{\infty}$, which diverges.

Answer 2

Hint: Add up two terms of the sum at a time. I.e., the $n=2k-1$ term and the $n=2k$ term, for $k=1,2,3,\ldots$.

You should find that the answer is asymptotically $1/k$.

Answer 3

$n=2m\implies\dfrac{-2}{4m}=-\dfrac1{2m}$

$n=2m-1\implies\dfrac4{2(2m-1}=\dfrac2{2m-1}$

If the number of terms is even,

$\sum_{m=1}^r\dfrac{2m+1}{2m(2m-1)}$

By comparison test, the series is divergent for $r\to\infty$

Can you handle the odd case?