Let X be a real Banach space and $(x_k^*)_{k\in N}$ be a sequence in $X^*$ with $\|x_k^*\| \leq 1$ for all $k \in N$. Show that the operator $T: X \rightarrow l^2(R) , Tx := \sum_{k=1}^{\infty} \frac{x_k^*(x)}{k!}e_k$ is compact, where $e_k$ denotes the unit vector from an orthonormal basis $l^2(R)$.
I only thought of making use of the following corollary but not quite sure: Let X,Y be Banach spaces and $T: X\rightarrow Y$ be linear and continuous, assume there exists a sequence $(T_n)_{n\in N}$ of linear continuous operators $T_n: X \rightarrow Y$ such that $\lim_{n\rightarrow \infty}\|T_n-T\| = 0$, dim($T_n(X))<\infty$ for all $ n \in N$. Then T is compact.
Let $T_{n}=\displaystyle\sum_{k=1}^{n}\dfrac{x_{k}^{\ast}(\cdot)}{k!}e_{k}$, then $T_{n}(X)\subseteq\left<e_{1},...,e_{n}\right>$ and $\|T_{n}\|\leq\displaystyle\sum_{k=1}^{n}\dfrac{1}{k!}$.
Note that $\|T(x)-T_{n}(x)\|\leq\displaystyle\sum_{k\geq n+1}\dfrac{|x_{k}^{\ast}(x)|}{k!}\|e_{k}\|\leq\sum_{k\geq n+1}\dfrac{\|x\|}{k!}$, so $\|T-T_{n}\|\leq\displaystyle\sum_{k\geq n+1}\dfrac{1}{k!}$.