Recall that an operator $T:X\rightarrow Y$ is called absolutely summing if there exists a constant $C>0$ such that $$\sum_{i=1}^{n}\|Tx_{i}\|\leq C \sup_{x^{*}\in B_{X^{*}}}\sum_{i=1}^{n}|\langle x^{*},x_{i}\rangle|,$$ for all finite families $(x_{i})_{i=1}^{n}$ in $X$. The least $C$ for which the above inequality always holds is denoted by $\pi(T)$. The set of all absolutely summing operators from $X$ to $Y$ is denoted by $\Pi(X,Y)$.

An operator $T:X\rightarrow Y$ is said to be nuclear provided that there are sequences $(x^{*}_{n})_{n}$ in $X^{*}$ and $(y_{n})_{n}$ in $Y$ such that $\sum\limits_{n}\|x^{*}_{n}\|\|y_{n}\|<\infty$ and $Tx=\sum\limits_{n}x^{*}_{n}(x)y_{n}$ for $x\in X$. The nuclear norm $\nu(T)$ is defined to be the infimum of $\sum\limits_{n}\|x^{*}_{n}\|\|y_{n}\|<\infty$ over all possible such representations. The space of all nuclear operators from $X$ to $Y$ is denoted by $\mathcal{N}(X,Y)$. It is known that every nuclear operator is absolutely summing.

Question 1. For $1<p,q<\infty$, when is $\Pi(l_{p},l_{q})$ reflexive ?

Question 2. For $1<p,q<\infty$, $\Pi(l_{p},l_{q})=\mathcal{N}(l_{p},l_{q})$ ?

Question 3. For $1<p<q<\infty$, is the inclusion map $i_{p,q}$ from $l_{p}$ to $l_{q}$ absolutely summing ?

Thank you !