Absolutely convergent function I am trying to show that if $\displaystyle\sum_{n\le x}f(n)=Cx+O(x^{3/4})$, where $f$ is non-negative multiplicative function and $C$ is a positive constant, then $F(s)=\displaystyle\sum_{n=1}^{\infty}f(n)n^{-s}$ is absolutely convergent for $\sigma>1$ and  $F_0(s)=\displaystyle\sum_{n=1}^{\infty}(f(n)-C)n^{-s}$ is convergent for $\sigma>3/4$. ($\sigma$ is the real part of complex number $s$).
As the guidance of @Fundamental (using partial summation), I have that
$\displaystyle\sum_{n=1}^{N}\dfrac{f(n)}{n^s}=\dfrac{CN+O(N^{3/4})}{N^s}+s\int_1^N\dfrac{Ct+O(t^{3/4})}{t^{s+1}}dt$ 
$=\dfrac{C}{N^{s-1}}+O(\dfrac{1}{N^{\sigma-3/4}})+s\displaystyle\int_1^N\dfrac{C}{t^s}+O(\dfrac{1}{t^{\sigma+1/4}})dt$  
$=\dfrac{C}{N^{s-1}}+O(\dfrac{1}{N^{\sigma-3/4}})+s[\dfrac{C}{(1-s)N^{s-1}}+O(\dfrac{1}{(3/4-\sigma)N^{\sigma-3/4}})+\dfrac{C}{s-1}+O(1)]$.
Since $\sigma>1$,Taking $N\rightarrow \infty$, we have $\displaystyle\sum_{n=1}^{\infty}\dfrac{f(n)}{n^s}=\dfrac{Cs}{s-1}+O(|s|)$.
However, according to the problem I have, I should have $\displaystyle\sum_{n=1}^{\infty}\dfrac{f(n)}{n^s}=\dfrac{C}{s-1}+O(|s|)$. Did I make some mistakes? Could anyone help me please? 
 A: First Problem
Define $A(0)=0$ and
$$
A(n)=\sum_{k=1}^nf(k)\tag{1}
$$
By hypothesis, we have
$$
A(n)=Cn+O\left(n^{3/4}\right)\tag{2}
$$
Furthermore, $f(n)=A(n)-A(n-1)$ and using the Mean Value Theorem,
$$
k^{-\sigma}-(k+1)^{-\sigma}\le\sigma k^{-\sigma-1}\tag{3}
$$
we have
$$
\begin{align}
\sum_{k=1}^\infty\left|f(k)k^{-s}\right|
&=\sum_{k=1}^\infty(A(k)-A(k-1))k^{-\mathrm{Re}(s)}\\
&=\lim_{n\to\infty}\sum_{k=1}^n\left(A(k)-A(k-1)\right)k^{-\mathrm{Re}(s)}\\
&=\lim_{n\to\infty}\left[\sum_{k=1}^nA(k)k^{-\mathrm{Re}(s)}-\sum_{k=1}^{n-1}A(k)(k+1)^{-\mathrm{Re}(s)}\right]\\
&=\lim_{n\to\infty}A(n)n^{-\mathrm{Re}(s)}+\sum_{k=1}^\infty A(k)\left(k^{-\mathrm{Re}(s)}-(k+1)^{-\mathrm{Re}(s)}\right)\tag{4}
\end{align}
$$
Thus, $(4)$ shows absolute convergence when
$$
\lim_{n\to\infty}A(n)\,n^{-\mathrm{Re}(s)}=0\tag{5}
$$
and
$$
\sum_{k=1}^\infty A(k)\,k^{-\mathrm{Re}(s)-1}\tag{6}
$$
converges.
Both $(5)$ and $(6)$ are satisfied because of $(2)$ when $\mathrm{Re}(s)\gt1$.

Second Problem
Define $A(0)=0$ and
$$
A(n)=\sum_{k=1}^n(f(k)-C)\tag{7}
$$
By hypothesis, we have
$$
A(n)=O\left(n^{3/4}\right)\tag{8}
$$
Furthermore, $f(n)-C=A(n)-A(n-1)$ and because
$$
\begin{align}
\left|k^{-s}-(k+1)^{-s}\right|
&=\left|\,\int_k^{k+1}s\,x^{-s-1}\,\mathrm{d}x\,\right|\\
&\le\int_k^{k+1}|s|\,x^{-\mathrm{Re}(s)-1}\,\mathrm{d}x\\[9pt]
&\le|s|k^{-\mathrm{Re}(s)-1}\tag{9}
\end{align}
$$
we have
$$
\begin{align}
\sum_{k=1}^\infty(f(k)-C)k^{-s}
&=\sum_{k=1}^\infty(A(k)-A(k-1))k^{-s}\\
&=\lim_{n\to\infty}\sum_{k=1}^n\left(A(k)-A(k-1)\right)k^{-s}\\
&=\lim_{n\to\infty}\left[\sum_{k=1}^nA(k)k^{-s}-\sum_{k=1}^{n-1}A(k)(k+1)^{-s}\right]\\
&=\lim_{n\to\infty}A(n)n^{-s}+\sum_{k=1}^\infty A(k)\left(k^{-s}-(k+1)^{-s}\right)\tag{10}
\end{align}
$$
Thus, $(10)$ shows convergence when
$$
\lim_{n\to\infty}A(n)\,n^{-\mathrm{Re}(s)}=0\tag{11}
$$
and
$$
\sum_{k=1}^\infty A(k)\,k^{-\mathrm{Re}(s)-1}\tag{12}
$$
converges. Both $(11)$ and $(12)$ are satisfied because of $(8)$ when $\mathrm{Re}(s)\gt\frac34$.
