Let $A$ be an operator $A: l_p \rightarrow l_p , 1 <p<\infty$ $$A(x_1, ..., x_n, ...)=\left(x_1, \frac{x_1+x_2}{2}, ..., \frac{x_1+...+x_n}{n}, ...\right)$$ I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=\left\{ \left(x_1, \frac{x_1+x_2}{2}, ..., \frac{x_1+...+x_n}{n}\right) : ||x_n||_{l_p}\le 1 \right\} $ is not relatively compact.
I have a criterion for that: I know that the subset $K \subset l_p $ is relatively compact iff $K$ is bounded and $\lim_{N\rightarrow\infty}\sup_ {x \in K}\sum_{n=N}^{\infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.
I would be grateful for any help!
Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1\leqslant i\leqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1\leqslant n\leqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies $$ (Av)(n)=2^{-(N+1)/p}\frac 1n\cdot n=2^{-(N+1)/p} $$ hence $$ \sum_{n=2^N+1}^{2^{N+1}}\left\lvert (Av)(n)\right\rvert^p=2^N\left(2^{-(N+1)/p}\right)^p=2^{-1}. $$ This proves, by the mentioned compactness criterion, that the set $\left\{Av^{(N)},N\geqslant 1\right\}$ is not relatively compact in $\ell^p$ hence that $A$ is not a compact operator.