= Solution
{c}
<absolute convergence> depends only on $\sigma=\operatorname{Re}s$, since $|a_nn^{-s}|=|a_n|n^{-\sigma}$. Let $A$ be the set of real $\sigma$ for which $\sum|a_n|n^{-\sigma}$ converges. Comparison with a convergent member shows that $A$ is upward closed. The hypotheses give a member of $A$ and a real number outside $A$, so its <infimum> $\sigma_0$ is finite. If $\sigma>\sigma_0$, some $\tau\in A$ lies below $\sigma$, giving convergence by comparison; if $\sigma<\sigma_0$, convergence would contradict the definition of the <infimum>. Thus \b[$\sigma_0=\inf A$ is the <abscissa of absolute convergence of a Dirichlet series>]. This argument leaves convergence on the boundary open.
Partition $1,\ldots,n$ according to $d=n/\gcd(k,n)$. Writing $k=(n/d)j$ identifies the part indexed by $d\mid n$ with the <integers> $1\leq j\leq d$ <coprime> to $d$. There are $\phi(d)$ of them. Counting the entire partition proves the <totient divisor-sum identity>
$$
\boxed{\sum_{d\mid n}\phi(d)=n.}
$$
We use the multiplication theorem for absolutely convergent <Dirichlet series>: their product is the <Dirichlet series> of the <Dirichlet convolution> of their coefficients. <absolute convergence> justifies rearrangement of the double sum. Here $0<\phi(n)\leq n$, so its series converges absolutely for $\sigma>2$, and
$$
\zeta(s)\sum_{n\geq1}\frac{\phi(n)}{n^s}
=\sum_{n\geq1}\frac{\sum_{d\mid n}\phi(d)}{n^s}
=\sum_{n\geq1}n^{1-s}=\zeta(s-1).
$$
The <Euler product> implies $\zeta(s)\neq0$ for $\sigma>1$. Division gives
$$
\boxed{\sum_{n\geq1}\frac{\phi(n)}{n^s}
=\frac{\zeta(s-1)}{\zeta(s)}\quad(\operatorname{Re}s>2).}
$$
We finally use the <simple pole> of the <Riemann zeta function> at one, of residue one, and $0<\zeta(2)<\infty$. The quotient therefore tends to $+\infty$ as real $s\downarrow2$. If $\sum\phi(n)n^{-2}$ converged, its nonnegative terms would dominate the series at every real $s>2$, giving a uniform finite bound and a contradiction. Thus the series diverges at two; comparison then excludes <absolute convergence> at every $\sigma<2$. Combined with the convergence already proved, \b[the abscissa is exactly two]. This is the <totient Dirichlet series and its absolute convergence boundary>.
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