I can understand that A) is true for
When odd-> n < 2^n, so as n-> infinity, 2^n grows more rapidly than n and it converges.. and same for even..
But is there any other proper method of solving these type of sums and find actually where( which point) the sequence converge to.
As you can see $|a_n|^{\frac{1}{n}}=n^{\frac{1}{n}}2^{-1} $ when $n$ is odd and $|a_n|^{\frac{1}{n}}= 3^{-1}$. Also $ \lim n^{\frac{1}{n}} =1 $ as n tends to $\infty$. Therefore $(B)$ is false as you can find two different subsequences which have different limits .But $\limsup |a_n|^{\frac{1}{n}}$ is strictly less than $1$ as $n \rightarrow \infty$. Therefore by root test $(D)$ is true. Thus $(C)$ is also true, as by $(D)$ the series is absolutely convergent.