Prove that $\sum_{n=1}^\infty (-1)^n \frac1n$ is convergent but not unconditionally convergent

Question

If $\left\{f_k\right\}_{k=1}^{\infty}$ is a sequence in a normed vectore space $X$ and $\sum_{k=1}^{\infty} f_{\sigma(k)}$ is convergent for all permutations $\sigma$, we say that $\sum_{k=1}^{\infty} f_k$ is unconditionally convergent.

I would like to prove that the following series $$\sum_{n=1}^\infty (-1)^n \frac1n$$ is not unconditionally convergent.

My attempt: $$\sum_{n=1}^\infty (-1)^n \frac1n=\sum_{k=1}^\infty (-1)^{2k} \frac{1}{2k}+\sum_{k=1}^\infty (-1)^{2k-1} \frac{1}{2k-1}=\sum_{k=1}^\infty \frac{1}{2k}-\sum_{k=1}^\infty \frac{1}{2k-1}$$ But $\sum_{k=1}^\infty \frac{1}{2k}$ is divergent.

P.S. This question follows my attempt to find a sequence $\left\{a_k\right\}_{k=1}^{\infty}$ of real numbers for which $\sum_{k=1}^{\infty} a_k$ is convergent but not unconditionally convergent, to understand the definition of unconditionally convergence.

Answer 1

For the convergence, you can use the Leibnitz Test. We consider $a_n=\frac{1}{n}$ . The $(a_n)$ is a decreasing sequence, and $\lim_{n\to \infty}a_n=0.$ Thus, we have by Leibnitz Test that the series $\sum_{n=0}^{\infty}(-1)^n\frac{1}{n}$ is convergent.

For the absolute convergence you can see that the series $\sum_{n=1}^{\infty}|a_n|$ does not converge.

Answer 2

$\lim_{N\rightarrow\infty}\sum_{n=1}^{N}(-1)^n\frac{1}{n}$ does exists because $$ (-1)^n\frac{1}{n}+(-1)^{n+1}\frac{1}{n+1} \\ = (-1)^n\left[\frac{1}{n}-\frac{1}{n+1}\right] \\ = (-1)^n\left[\frac{1}{n(n+1)}\right] $$ However, the series is not absolutely convergent.