a linear map on $W$ 
Define $W = \{(a_1, a_2,\cdots) : a_i \in \mathbb{F}, \exists N\in\mathbb{N}, \forall n \geq N, a_n = 0\},$ where $\mathbb{F} = \mathbb{R} $ or $\mathbb{C}$ and $W$ has the standard inner product, which is given by $\langle(a_1,a_2,\cdots), (b_1,b_2,\cdots)\rangle = \sum_{i=1}^\infty a_i \overline{b_i}$ ($\overline{b_i}$ is simply the complex conjugate of $b_i$). Prove that the linear map $T : W \to W$ given by $T(a)_j = \sum_{i=j}^\infty a_i$, where $T(a) = (T(a)_1, T(a)_2,\cdots),$ has no adjoint.

I know that to show that $T$ has an adjoint $T^*$, it suffices to show that for all $a,b \in W$, $\langle T(a),b \rangle = \langle a, T^* b\rangle$. So to show that $T$ does not have an adjoint, it suffices to show that there is no linear map $T^*$ so that for all $a,b \in W \langle T(a),b\rangle = \langle a,T^*b\rangle$ For any $a \in W,$ we may find $N\in\mathbb{N}$ st $i \geq N\Rightarrow a_i = 0.$ Hence
$$\langle T(a),b\rangle = \sum_{i=1}^\infty \left(\sum_{j=i}^\infty a_j\right) \overline{b_i}=\sum_{i=1}^{N-1} \left(\sum_{j=i}^{N-1} a_j\right)\overline{b_i}.$$
Also,
$$\langle a, T^*b\rangle = \sum_{i=1}^\infty a_i\overline{(T^*b)_i}=\sum_{i=1}^{N-1} a_i\overline{(T^*b)_i}.$$
Hence
$$\langle T(a),b\rangle = \langle a,T^*b\rangle \iff \sum_{i=1}^{N-1} \left(\sum_{j=1}^{N-1} a_j \overline{b_i}-a_i \overline{(T^*b)_i}\right) = 0.$$
I know I'm supposed to find a $b \in W$ that'll make it impossible for $\langle a,T^*b\rangle = \langle T(a),b\rangle$ for all $a\in W$, but I'm unsure how to find this.

Let $e_i\in W$ satisfy $(e_i)_j=1$ if $i=j$ and $(e_i)_j=0$ otherwise.  Then $$\langle Te_i,e_j\rangle=\cases{1\,\,{\rm if}\,\,j\leq i,\\0 \,\,{\rm if}\,\,j>i.}$$
Thus if $T^*$ exists: $$\langle e_i,T^*e_j\rangle=\cases{1\,\,{\rm if}\,\,j\leq i,\\0 \,\,{\rm if}\,\,j>i.}$$
So we have $(T^*e_j)_i=1$ for all $i\geq j$ which contradicts $T^*e_j\in W$.
