What is $\lim \sup_{n \to \infty} \sum_{k=1}^{n} (-1)^k \left(\frac{1}{2} -\frac{1}{k} \right)$?

Question

Let $a_{n}=(-1)^n \left( \dfrac{1}{2} -\dfrac{1}{n}\right)$ and let $b_n=\sum\limits_{k=1}^{n} a_{k}, \forall \ n \in \mathbb{N}$.

Then what is $\lim \sup b_{n}$ and $\lim \inf b_{n}$?

$\begin{align} \lim_{n \to \infty} b_{n} &= \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{2} -\frac{1}{n}\right)\\ &= \sum_{n=1}^{ \infty} (-1)^n \frac{1}{2} -\sum_{n=1}^{\infty}(-1)^n \frac{1}{n} \\ &=\sum_{n=1}^{\infty}(-1)^n \frac{1}{2} -\log 2 \end{align}$

So $\lim \inf b_{n}= \frac{-1}{2}-\log 2$ and $\lim \sup b_{n}= -\log 2$

Is it correct?

Answer 1

Seems like you committed a small mistake

\begin{align} \lim_{n \to \infty} b_{n} &= \sum_{n=1}^{\infty} (-1)^n \big( \dfrac{1}{2} -\dfrac{1}{n}\big)\\ &= \sum_{n=1}^{ \infty} (-1)^n \dfrac{1}{2} -\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n} \\ &=\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{2}+\log 2 \end{align}

Which implies that as

$b_{2n}=\log 2+\frac{1}{2}$

$b_{2n+1}=\log 2-\frac{1}{2}$

$\lim \inf b_{n}=\log 2- \frac{1}{2}$

$\lim \sup b_{n}=\log 2 + \frac{1}{2}$