Consider a sequence $\{a_n\}_n$,
$$ a_n=(-1)^n\left(\frac{1}{2}-\frac{1}{n}\right).$$
Let $b_n=\sum_{k=1}^n a_k$, $\forall\; n\in\mathbb{N}$. Then which of the following is true?(i) $\lim_{n\to\infty} b_n=0$
(ii) $\lim\sup_{n\to\infty} b_n>\frac{1}{2}$
(iii) $\lim\inf_{n\to\infty} b_n<-\frac{1}{2}$
(iv) $0\le\lim\inf_{n\to\infty} b_n\le\lim\sup_{n\to\infty} b_n\le\frac{1}{2}$
Now $\lim{a_{2n-1}}=-\frac{1}{2}$ and $\lim{a_{2n}}=\frac{1}{2}$. So my intuition says that (i) might be true. But I have got no such theorem that supports my thinking.
Again $\lim{a_{2n-1}}=-\frac{1}{2}$ may indicate that $\lim\inf_{n\to\infty} b_n\ge-\frac{1}{2}$ and $\lim{a_{2n}}=\frac{1}{2}$ may indicate that $\lim\sup_{n\to\infty} b_n\le\frac{1}{2}$. So (ii) and (iii) is not possible.
(iv) could be a possibility.
I am unable to solve this problem. But MSE always wants a try from the poster. So I posted this although I know that there are some mistakes.
SOURCE: CSIR-UGC NET
Best Answer
Asserting that $\lim_{n\to\infty}b_n=0$ is the same thing as asserting that $\sum_{n=1}^\infty a_n=0$. But the series $\sum_{n=1}^\infty a_n$ diverges, since you don't have $\lim_{n\to\infty}a_n=0$.
On the other and, if $n\in\Bbb N$,\begin{align}b_n&=\sum_{k=1}^n(-1)^k\left(\frac12-\frac1k\right)\\&=\left(\sum_{k=1}^n\frac{(-1)^k}2\right)+\left(\sum_{k=1}^n\frac{(-1)^{k-1}}k\right),\end{align}and$$\sum_{k=1}^n\frac{(-1)^k}2=\begin{cases}-\frac12&\text{ if $n$ is odd}\\0&\text{otherwise.}\end{cases}$$On the other hand, for each $n\in\Bbb N$,$$\sum_{k=1}^n\frac{(-1)^{k-1}}k\in\left[\frac12,1\right],$$since, if $(x_n)_{n\in\Bbb N}$ is a decreasing sequence of real numbers, you always have$$\sum_{k=1}^n(-1)^{k-1}x_k\in[x_1-x_2,x_1].$$Therefore, the second option is the correct one, since$$\lim_{n\to\infty}b_{2n}=\sum_{n=1}^\infty\frac{(-1)^{n-1}}n=\log(2)>\frac12.$$