Adjoint operator $T^{\ast}$ in space $l^1$ 
Find the adjoint operator of $T:l^1 \rightarrow l^1$, $(x_k)_{k\in \mathbb{N}}\mapsto (\sum^{\infty}_{k=1}x_k,0,0,...)$

In our lecture we defined the adjoint operator as
\begin{align*}
T^{\ast}(y^{\ast}):X\rightarrow \mathbb{K}\\
x\mapsto y^{\ast}(Tx)
\end{align*}
so I think I must do something like this...
\begin{align*}
(T^{\ast}y^{\ast})(x) = y^{\ast}(Tx) = y^{\ast}((\sum^{\infty}_{k=1}
x_k,0,0,...))= y_1^{\ast}(\sum^{\infty}_{k=1} x_k)+y_2^{\ast}(0)+y_3^{\ast}(0)+...=y_1^{\ast}(\sum^{\infty}_{k=1} x_k)=y_1^{\ast}(x_1)+y_1^{\ast}(x_2)+...
\end{align*}
Because $y^{\ast}_j$ is linear $\forall j\in \mathbb{N}$ $\Rightarrow y_j^{\ast}(0)=0$
So it appears to me that my $T^{\ast}$ must satisfy $T^{\ast}(y_1,y_2,y_3,...)=(y_1,y_1,y_1,...)$.
Is that correct? Can I write my $T^{\ast}$ more explicit?
 A: What you are writing here is correct, well done! I am adding the details behind the calculations:
The adjoint of an operator $T:X\to Y$ between Banach spaces is the operator $T^*:Y^*\to X^*$ defined by $T^*(\phi)=\phi\circ T$. It is easy to see that $T^*$ is a well-defined linear operator and $\|T^*\|=\|T\|$ so when $T$ is bounded, $T^*$ is bounded.
In our case, we have that $T^*:(\ell^1)^*\to(\ell^1)^*$ defined as $T^*(\phi)=\phi\circ T$. Now assume that $(x_n)\in\ell^1$. Then
$$T^*(\phi)(x_n)=\phi\circ T(x_n)=\phi(\sum_{k=1}^\infty x_k,0,0,\dots)$$
We have an isometric isomorphism $(\ell^1)^*\cong\ell^\infty$ which is defined by $J:\ell^\infty\to(\ell^1)^*$ as follows: if $y=(y_n)\in\ell^\infty$, then $y\mapsto J_y$, where $J_y(x_n)=\sum_ny_nx_n$.
Now for $y=(y_n)\in\ell^\infty$ we have that
$$T^*(J_y)(x_n)=J_y(\sum_{k=1}^\infty x_k,0,0,\dots)=y_1\sum_{k=1}^\infty x_k+y_2\cdot0+y_3\cdot0+\dots=\sum_{k=1}^\infty y_1x_k=J_{z}(x_n)$$
where $z=(y_1,y_1,y_1,\dots)\in\ell^\infty$. This shows that $T^*(J_y)=J_z$, i.e. if we identify $(\ell^1)^*$ with $\ell^\infty$ through the isometric isomorphism $J$, then $T^*(y_1,y_2,y_3,\dots)=(y_1,y_1,y_1,\dots)$, as OP states.
