= Solution
For $N=\lfloor x\rfloor$, interchange a finite sum and an integral:
$$
\int_1^x S(t)F'(t)\,dt
=\sum_{n\le x}f(n)\int_n^x F'(t)\,dt
=F(x)S(x)-\sum_{n\le x}f(n)F(n).
$$
This proves the <Abel's summation formula>, including the term at $n=1$.
Apply it with $f(n)=1$, $F(t)=1/t$. Writing $\{t\}=t-\lfloor t\rfloor$ gives
$$
H_{\lfloor x\rfloor}=\log x+1-\frac{\{x\}}x-\int_1^x\frac{\{t\}}{t^2}\,dt
=\log x+\gamma-\frac{\{x\}}x+\int_x^\infty\frac{\{t\}}{t^2}\,dt.
$$
The last two terms are $O(x^{-1})$, proving the first expansion. For the refinement let $B_1(t)=\{t\}-1/2$. It has mean <zero> on each unit interval, so its <primitive> $P(t)=\int_0^tB_1(u)\,du$ is bounded and periodic. <Integration by parts> gives
$$
\int_x^\infty\frac{B_1(t)}{t^2}\,dt
=-\frac{P(x)}{x^2}+2\int_x^\infty\frac{P(t)}{t^3}\,dt=O(x^{-2}).
$$
Thus for <integer> $x$, where $\{x\}=0$,
$$
\boxed{H_x=\log x+\gamma+\frac1{2x}+O(x^{-2}).}
$$
For the zeta representation, integrating each defining integral by parts yields
$$
sf_n(s)=-(n+1)^{-s}+\frac{(n+1)^{1-s}-n^{1-s}}{1-s}.
$$
Summing from $1$ to $N-1$ and substituting into the definition of $\zeta$ gives
$$
\boxed{\zeta(s)=\frac{N^{1-s}}{s-1}+\sum_{n\le N}n^{-s}-s\sum_{n=N}^\infty f_n(s).}
$$
Since $0\le x-n\le1$,
$$
\left|\sum_{n=N}^\infty f_n(s)\right|
\le\sum_{n=N}^\infty\int_n^{n+1}x^{-\sigma-1}\,dx
=\frac{N^{-\sigma}}\sigma.
$$
This is uniform on compact subsets of $\operatorname{Re}s>0$, so the series is <holomorphic> there and limits at $s=1$ are legitimate after subtracting the <pole>. Expanding $N^{1-s}=1-(s-1)\log N+O((s-1)^2)$ gives
$$
a_0=H_N-\log N-\sum_{n=N}^\infty f_n(1).
$$
The final tail is $O(N^{-1})$, and the harmonic expansion sends the remaining difference to $\gamma$. Therefore
$$
\boxed{a_0=\gamma.}
$$
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