= Solution
The <continuous dual space> $X'$ consists of all bounded complex-linear <functionals> $f:X\to\mathbb C$, with pointwise vector operations and <norm> $\|f\|=\sup_{\|x\|\leq1}|f(x)|$. This is finite by boundedness. The scalar <norm> identity and <triangle inequality> follow under the supremum; if the <norm> is zero, scaling every nonzero vector into the unit ball shows that $f$ vanishes. Hence $X'$ is a normed <complex vector space>.
For the <dual of the convergent sequence space>, take $\beta\in\ell^1$. The displayed formula is a <bounded linear functional>, since the <series> is absolutely convergent and
$$
|\phi(\beta)(\alpha)|\leq\left(|\beta_1|+\sum_{j\geq1}|\beta_{j+1}|\right)\|\alpha\|_\infty.
$$
It is linear in $\beta$, so $\|\phi(\beta)\|\leq\|\beta\|_1$. For equality fix $N$, choose $\alpha_j$ for $j\leq N$ of modulus at most one so that $\alpha_j\beta_{j+1}=|\beta_{j+1}|$, and make all later terms equal a constant $c$ of modulus at most one with $c\beta_1=|\beta_1|$. The choices for a zero coefficient can be arbitrary. This sequence lies in the unit ball of $c$ and converges to $c$. The <reverse triangle inequality> gives
$$
|\phi(\beta)(\alpha)|\geq|\beta_1|+\sum_{j=1}^N|\beta_{j+1}|-\sum_{j>N}|\beta_{j+1}|.
$$
Letting $N\to\infty$ proves \b[$\|\phi(\beta)\|=\|\beta\|_1$], including injectivity.
For surjectivity let $F\in c'$. By the permitted description of $c_0'$, there is $\gamma\in\ell^1$ with $F(z)=\sum_j\gamma_jz_j$ on $c_0$. Every $\alpha\in c$ decomposes as $\alpha-\alpha_\infty\mathbf1\in c_0$ plus $\alpha_\infty\mathbf1$. Consequently
$$
F(\alpha)=\sum_j\gamma_j\alpha_j+\alpha_\infty\left(F(\mathbf1)-\sum_j\gamma_j\right).
$$
Take $\beta_{j+1}=\gamma_j$ and $\beta_1=F(\mathbf1)-\sum_j\gamma_j$. This is an $\ell^1$ sequence and $F=\phi(\beta)$, proving \b[the stated isometric linear isomorphism onto $c'$]. No complex conjugation is inserted, since these are complex-linear dual <functionals>.
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