$\|x\|_{p} \rightarrow\|x\|_{\infty}$ as $p \rightarrow \infty$ The following question came in my quiz:

(True/False) Let $x \in \ell^{p_{0}}$ for some $1 \leq p_{0}<\infty .$ Then $\|x\|_{p} \rightarrow\|x\|_{\infty}$ as $p \rightarrow \infty$.

Where $\ell^{p}=\left\{x=\left(x_{1}, x_{2}, \ldots\right): \sum_{j=1}^{\infty}\left|x_{j}\right|^{p}<\infty\right\}$ for $1 \leq p<\infty$,
$
\|x\|_{p}=\left(\sum_{j=1}^{\infty}\left|x_{j}\right|^{p}\right)^{\frac{1}{p}}$ and $\|x\|_{\infty}=\sup_{j \in \mathbb{N}}\left|x_{j}\right|$. I answered it true and gave the following reason:
Let $x \in \ell^{p_{0}}$ for some $1 \le p_{0}<\infty$, then $x \in \ell^{p}$ $\forall\, p \ge p_{0}$ since $\ell^{p_{0}} \subset \ell^{p}$. This implies $\sum_{j=1}^{\infty}\left|x_{j}\right|^{p}$ is finite $\forall\, p \ge p_{0}$. If $\|x\|_{\infty}=0$, then $x=0$ and the convergence is trivial. If $\|x\|_{\infty}\ne0$, then $$\tag{1} \|x\|_{p}=\|x\|_{\infty}\left(\sum_{j=1}^{\infty}\left(\frac{\left|x_{j}\right|}{\|x\|_{\infty}}\right)^{p}\right)^{1 / p}.$$
Since $\lim _{j \rightarrow \infty}\left|x_{j}\right|=0$, the sequence $\left(\left|x_{n}\right|\right)_{n=1}^{\infty}$ converges to zero and $ \sup _{j \in \mathbb{N}}\left\{\left|x_{j}\right|\right\}=\left|x_{j^{\prime}}\right|$ for some $j^{\prime} \in \mathbb{N}$. Let $J:=\left\{j \in \mathbb{N}:| x_{j}|=\|x\|_{\infty}\right\}$, then $J$ is non empty $\left(j^{\prime} \in J\right)$ and the cardinality of $J$ is finite (otherwise $\left.\|x\|_{\infty}=0\right)$. Therefore, $ \displaystyle\lim_{p \to \infty} \sum_{j=1}^{\infty} \frac{\left|x_{j}\right|^{p}}{||x||_{\infty}^{p}} = |J|+ \lim_{p \to \infty}\sum_{j \notin J}\frac{\left|x_{j}\right|^{p}}{||x||_{\infty}^{p}}$ and since $\sum\left|x_{j}\right|^{p}$ is finite $\forall\, p \ge p_{0}$, $\sum \frac{\left|x_{j}\right|^{p}}{||x||_{\infty}^{p}}$ is also finite. Thus $$\tag{2} \lim_{p \to \infty} \sum_{ j\notin J} \frac{\left|x_{j}\right|^{p}}{||x||_{\infty}^{p}} = \sum_{j \notin J} \lim_{p \to \infty} \frac{\left|x_{j}\right|^{p}}{||x||_{\infty}^{p}} = 0 $$ and the result follows from $(1)$, as $||x||_{p} = ||x||_{\infty} $exp$\left( \frac{1}{p} \ln{\sum_{j=1}^{\infty} \frac{\left|x_{j}\right|^{p}}{||x||_{\infty}^{p}}}\right)$
However, the explanation in the answer sheet is marked incorrect. I don't understand where my mistake it.
 A: Yes, that should be true. As you have stated if $x\in\mathcal l_1$ then $|x_k|\to 0$. Thus $|x_k|$ takes a maximum. WLOG we can assume that this maximum is $1$ (else divide by this maximum, as for any norm we have $||x/c||=||x||/|c|$). Thus $|x_k|\leq 1$.
Then:
$$ 1 \leq ||x||_p^p = \sum_{k} |x_k|^p \leq \sum_{k} 1^{p-1}|x_k| = ||x||_1 $$
Thus
$$ 1\leq ||x||_p \leq \sqrt[p]{||x||_1} \to 1$$
A: In general if $\{y_n(p)\}$ is a sequence (depending on $p$), then
$$ \lim_{p\to \infty} \sum_{n=1}^\infty y_n(p) = \sum_{n=1} ^\infty \lim_{p\to \infty} y_n(p)$$
might not hold. For example, if $y_n(p) = 1$ when $p<n$ and $0$ otherwise, then
$$+\infty =  \sum_{n=1}^\infty y_n(p) \neq \sum_{n=1}^\infty \lim_{p\to \infty } y_n(p) = 0$$
But in your case, (2) actually holds. However you would need an argument for it (e.g. it follows from monotone convergence theorem).
However you don't need (2). Instead, we have (for all $p\ge p_0$, where $p_0$ is fixed)
$$ |J|^{1/p} \le \left( |J| + \sum_{n\notin J} \left(\frac{\left|x_{j}\right|}{\|x\|_{\infty}}\right)^{p}\right)^{1/p} \le \left( |J| + \sum_{n\notin J}\left(\frac{\left|x_{j}\right|}{\|x\|_{\infty}}\right)^{p_0}\right)^{1/p} $$
taking $p\to\infty$ and use squeeze theorem, we have
$$\lim_{p\to \infty}  \left( |J| + \sum_{n\notin J} \left(\frac{\left|x_{j}\right|}{\|x\|_{\infty}}\right)^{p}\right)^{1/p} =1.$$
