Consider the sequence $\{a_n\},$ $$a_n=\left(-1\right)^n\left(\frac{1}{2}-\frac{1}{n}\right).$$ Let $$b_n=\sum _{k=1}^na_n,\forall n\in \mathbb N.$$
(A) $ \lim_{n \to \infty}b_n=0$
(B)$\limsup_{n\to \infty}b_n>1/2$
(C)$\liminf_{n\to \infty}b_n<-1/2$
(D)$0\leq \liminf_{n\to \infty}b_n \leq \limsup_{n\to \infty}b_n \leq 1/2$
My effort: $\lim_{n \to \infty}a_n\neq 0,$ So, $b_n=\sum _{k=1}^na_n,\forall n\in \mathbb N$ doesn't converge. So, (A) and (D) are false.
$b_1=1/2>1$. Assume $b_k>0.$ We need to prove $b_{k+1}=\sum _{j=1}^{k+1}a_j=b_k+\left(-1\right)^{k+1}\left(\frac{1}{2}-\frac{1}{k+1}\right).$ If I could complete the proof using induction, I could Eliminate (C). How do I complete the proof?
$$b_{2k} = \sum_{i=1}^{2k} (-1)^{n-1}\frac{1}{i} > \frac{1}{2}, \forall k \ge 2.$$ $$b_{2k+1} =\sum_{i=1}^{2k+1} (-1)^{n-1}\frac{1}{i} - \frac{1}{2} \ge 0.$$