Prove that $\sum |x_k| < \infty \iff \sum |x_k|^p < \infty$

Question

I'm trying to solve this problem about series. Could you please verify if my attempt is fine or it contains logical gaps/mistakes? I'm grateful for your help!

Let $\mathbb K$ denote $\mathbb R$ or $\mathbb C$, $(x_k)$ be a sequence in $\mathbb K$, and $1 \le p \in \mathbb R$. Then $$\sum |x_k| < \infty \iff \sum |x_k|^p < \infty$$

My attempt:

We define partial sums $s_n,s'_n$ by $$s_n = \sum_{k=0}^n |x_k|, \quad s'_n = \sum_{k=0}^n |x_k|^p$$

Lemma: For all $n \in \mathbb N$, we have $$(s_n)^p \ge s'_n \ge s_n$$

If $\sum |x_k| < \infty$ then there exists $M \ge 0$ such that $s_n \le M$ for all $n \in \mathbb N$. By Lemma, $s'_n \le (s_n)^p \le M^p$ for all $n \in \mathbb N$. As such, $s'_n$ is bounded from above and thus $\sum |x_k|^p < \infty$.

If $\sum |x_k|^p < \infty$ then there exists $N \ge 0$ such that $s'_n \le N$ for all $n \in \mathbb N$. By Lemma, $s_n \le s'_n \le L$ for all $n \in \mathbb N$. As such, $s_n$ is bounded from above and thus $\sum |x_k| < \infty$.

Answer 1

The implication is obvious since $|x_k|^p\leq |x_k|$ for all $k$ big enough but the converse is not true as $|x_k|=\frac{1}{k}$ prove.

Answer 2

Your lemma is false. For a correct proof of $\implies$ just note that $|x_k| <1$ for $k$ sufficiently large and $|x| <1$ implies $|x|^{p} \leq |x|$. The reverse implication is false since $\sum \frac 1 {n^{2}} <\infty$ but $\sum \frac 1 n =\infty$.