I want to check that my approach for this question makes sense. Consider the bounded linear map $$T:\ell_1 \to c_0$$ where $$c_0$$ is the space of sequences with limit zero, defined as $$T((c_n)_n) = (c_1 + c_2 +c_3 +\dots, c_2 + c_3 + \dots, c_3 + \dots, \dots)$$ for any $$(c_n)_n \in c_0$$. I want to find its adjoint $$T^*$$. Since $$c_0^* = \ell_1$$ and $$\ell_1^* = \ell_\infty$$, the adjoint is the map: $$T^*:\ell_1 \to \ell_{\infty}$$ with the following property:

Let $$(x_n) \in \ell_1$$ and $$(c_n) \in c_0$$ then

\begin{align*} [T^*((x_n)) ] (c_n) = (x_n) [T((c_n))] \end{align*} expanding the right hand side gives:

\begin{align*} &=(x_n) [(c_1 + c_2 +c_3 +\dots, c_2 + c_3 + \dots, c_3 + \dots, \dots)]\\ &= (x_n) \left ( \sum_{j=n}^{\infty}c_j \right )_{n=1}^{\infty}\\ &=\sum_{n=1}^{\infty} x_n \sum_{j=n}^{\infty}c_j\\ &= \sum_{n=1}^{\infty} (n c_n)x_n \end{align*}

So $$T^*$$ is given explicitly by the action of the sequence $$(n c_n)_n \in \ell_{\infty}$$ on elements of $$\ell_1$$.

Everything is perfect until the last equality, where you suddenly decided that $$\sum_{j=n}^\infty c_j=nc_n$$.
When you have $$\sum_{n=1}^{\infty} x_n \sum_{j=n}^{\infty}c_j,$$ you have $$n\geq1$$, $$j\geq n$$. So to switch the sums now you have $$j\geq1$$, $$1\leq n\leq j$$. Thus $$\sum_{n=1}^{\infty} x_n \sum_{j=n}^{\infty}c_j=\sum_{j=1}^\infty c_j\left(\sum_{n=1}^j x_n\right).$$ That shows that (using $$k$$ in place of $$n$$ and $$n$$ in place of $$j$$ above) $$T^*x=\left(\sum_{k=1}^n x_k\right)_n$$