Let $\left( x^{(n)}\right)_{n=1}^{\infty} \subset \ell^p$ be a Cauchy sequence. Since I see you have troubles with your notations of sequence of sequences, this is the notation that I will use for each element $x^{(n)}$ in the sequence:
$$
x^{(n)} = \left( x_j^{(n)}\right)_{j=1}^{\infty} = \left( x_1^{(n)},x_2^{(n)}, \cdots \right)\in \ell^p
$$
For $x= \left( x_j\right)_{j=1}^{\infty} , y= \left( y_j\right)_{j=1}^{\infty} \in \ell^p$, lets define the $p$-norm $\| \cdot \|_p$ as the one who induces $d_p$, that is $\|x-y\|_p=d_p(x,y)$. Precisely
$$
\|x-y\|_p= \left(\sum_{j=1}^{\infty} \left|x_j-y_j\right|^p\right)^{1/p}
$$
Now lets continue, take $\varepsilon>0$, then there exist a $N=N(\varepsilon) \in \mathbb{N}$, such that if $m,n >N$ then
$$
\|x^{(m)}-x^{(n)}\|_p<\varepsilon.
$$
Thus, for any $j \in \mathbb{N}$, it follows that
$$
\left|x^{(m)}_j-x^{(n)}_j\right|^p \leq \sum_{j=1}^{\infty} \left|x^{(m)}_j-x^{(n)}_j\right|^p = \|x^{(m)}-x^{(n)}\|^p_p<\varepsilon^p
$$
that is, for any $j \in \mathbb{N}$ the sequence $\left( x^{(n)}_j\right)_{n=1}^{\infty} \subset \mathbb{R}$ is a Cauchy one. Since $\mathbb{R}$ is complete, for each $j$ there exist a $x_j \in \mathbb{R}$ such that
$$
\lim_{n \to \infty} x^{(n)}_j = x_j
$$
Lets fix $k \in \mathbb{N}$, then in a similar way for $m,n >N$
\begin{equation}
\sum_{j=1}^{k} \left|x^{(m)}_j-x^{(n)}_j\right|^p \leq \sum_{j=1}^{\infty} \left|x^{(m)}_j-x^{(n)}_j\right|^p = \|x^{(m)}-x^{(n)}\|^p_p<\varepsilon^p \tag{1}
\end{equation}
Letting $n \to \infty$ in (1), we get that for $m>N$
\begin{equation}
\sum_{j=1}^{k}\left|x^{(m)}_j-x_j\right|^p < \varepsilon^p \tag{2}
\end{equation}
Then by the usual triangle inecuality ( Minkowski's inequality for $\|\cdot\|_p$ in $\mathbb{R}^k$) we get that if $m>N$
$$
\left( \sum_{j=1}^{k}|x_j|^p \right)^{1/p} \leq \left( \sum_{j=1}^{k}\left|x^{(m)}_j-x_j\right|^p \right)^{1/p} + \left( \sum_{j=1}^{k} \left|x^{(m)}_j \right| \right)^{1/p} < \varepsilon + \left( \sum_{j=1}^{k} \left|x^{(m)}_j \right| \right)^{1/p}
$$
by letting $k \to \infty$, we get $\|x\|_p\leq \varepsilon + \|x^{(m)}\|_p$, which is the same as getting that $x=\left( x_j\right)_{j=1}^{\infty} \in \ell^p$. Again, letting $k \to \infty$ in (2), we obtain that if $m>N$
$$
\|x^{(m)}-x\|_p^p= \sum_{j=1}^{\infty}\left|x^{(m)}_j-x_j\right|^p < \varepsilon^p
$$
thus
$$
\lim_{m \to \infty} \|x^{(m)}-x\|_p= 0
$$
so indeed, $\left( x^{(m)}\right)_{m=1}^{\infty} \subset \ell^p$, is a convergent sequence who converges to $x \in \ell^p$. We conclude then that $\ell^p$ is a complete metric space for $1\leq p < \infty$.
Hint: Let $(x_n)=\frac 1 N(1,1,..,1,0,0,...)$ where there are $N$ $1$'s. Then $\|A(x_n)\|=\frac 1 N \sum\limits_{k=1}^{\infty } \frac {a^{k}} {k!}(N-k) =\sum\limits_{k=1}^{\infty }\frac {a^{k}} {k!}(\frac {N-k} N )$. An Application of DCT shows that $\|A(x_n)\| \to e^{a}$ as $N \to \infty$. Hence $\|A\|=e^{a}$.
Best Answer
What you want to prove is what is sometimes called Jensen inequality, and even if only finitely many terms are nonzero it is not completely trivial.
Here is how a possible proof goes: You can divide the proof into 2 steps.
For step 1: This can be shown similarly as you did: Note that the proof does not run into your above two difficulties because
For step 2: Let $x\in\ell_p$. In case $x=0$, the inequality is trivial. In case $x\ne0$, there is a number $\lambda>0$ such that $y=\lambda x\in\ell_p$ satisfies $\lVert y\rVert_p=1$. Now apply step 1 to obtain that $\lVert y\rVert_q\le1=\lVert y\rVert_p$. Dividing this inequaliy by $\lambda>0$ gives the result.