Deduce the value $\sum_{n=1}^\infty f(n)n^{-s} $. Let $s_1 = \sigma_1 + it_1$ where $\sigma_1 > 0$. Set $F(x) = \sum_{n \leq x} f(n)$. Suppose that $$\sum_{n =1}^\infty \frac{f(n)}{n^{s_1}}$$ converge.
Deduce that $$F(x) = o(x^{\sigma_1}).$$ Then show that $$\int_{1}^\infty t^{-s_1 -1}F(t) dt$$ converges and $$\sum_{n=1}^\infty \frac{f(n)}{n^{s_1}} = s_1\int_{1}^\infty t^{-s_1 -1}F(t) dt.$$
Sol. : Set $a_n = \frac{f(n)}{n^{s_1}}$, $g(x) = x^{s_1}$ and $A(x) = \sum_{n \leq x} a_n.$ By partial summation, $$\sum_{n\leq x}f(n) = \sum_{n \leq x}\frac{f(n)}{n^{s_1}}n^{s_1}= A(x)x^{s_1} - s_1 \int_{1}^x t^{s_1 - 1}A(t) dt.$$
Since $\lim_{x \rightarrow \infty} A(x)$ exist, $A(x) = O(1).$ So $$\sum_{n\leq x}f(n) = O(x^{s_1}) + O(s_1\int_1^x t^{s_1-1} dt) = O(x^{s_1}).$$
So $$\frac{\sum_{n\leq x}f(n)}{x^{\sigma_1}} = O(1)$$ which does not yield $o(x^{\sigma_1})$. I think I go to rough on $A(x) = O(1)$, but not sure what I can say more about it.
$\textbf{Any help to fix the error please ?}$

Now suppose that $F(x) = o(x^{\sigma_1}).$ Set $$b_n = f(n), g(x) = x^{-s_1}, B(x) = \sum_{n \leq x} b_n.$$ By partial summation,
$$\sum_{n\leq x}f(n)n^{-s_1} = B(x)x^{-s_1} + s_1 \int_{1}^x t^{-s_1-1}B(t)dt = \frac{o(x^{\sigma_1})}{x^{s_1}} + s_1 \int_{1}^x t^{-s_1-1}o(t^{\sigma_1})dt.$$
Since $$\frac{o{x^{\sigma_1}}}{x^{s_1}} = o(1),$$ taking $x \rightarrow \infty$ then $$\sum_{n=1}^\infty f(n)n^{-s_1} = s_1 \lim_{x \rightarrow \infty} \int_1^x o(t^{-it-1})dt.$$
$\textbf{Not sure how to do with the last integral}$
 A: 
Any help to fix the error please?

Once you have replaced a term with $O(x^{\sigma_1})$, there's no hope to reach an $o(x^{\sigma_1})$ from there. You need to go back and try something else. A trick that often works is to add a suitable version of $0$. Here, we use
$$s_1 \int_1^x t^{s_1-1}\,dt = \bigl[t^{s_1}\bigr]_1^x = x^{s_1} - 1$$
to add $L x^{s_1} - L x{s_1}$, where $L = \sum_{n = 1}^\infty f(n) n^{-s_1}$:
\begin{align}
\sum_{n \leqslant x} f(n) &= A(x)x^{s_1} - s_1 \int_1^x t^{s_1-1}A(t)\,dt\\
&= \bigl(A(x) -L\bigr)x^{s_1} + L - s_1\int_1^x t^{s_1-1}\bigl(A(t) - L\bigr)\,dt.
\end{align}
Now we note that $A(x) - L \in o(1)$, so the first term in the sum belongs to $o(x^{\sigma_1})$. Since $\sigma_1 > 0$, we have $\lim_{x\to+\infty} x^{\sigma_1} = +\infty$, and hence the constant $L$ also belongs to $o(x^{\sigma_1})$. Finally, to deal with the last term, the integral, for $\varepsilon > 0$ choose an $x_{\varepsilon}$ such that $\lvert A(t) - L\rvert < \varepsilon$ for $t \geqslant x_{\varepsilon}$, and split the integral at $x_{\varepsilon}$. We find
$$\biggl\lvert s_1 \int_1^x t^{s_1-1}\bigl(A(t) -L\bigr)\,dt\biggr\rvert \leqslant \biggl\lvert s_1\int_1^{x_\varepsilon} t^{s_1-1}\bigl(A(t)-L\bigr)\,dt\biggr\rvert + \varepsilon \bigl(x^{\sigma_1} - x_\varepsilon^{\sigma_1}\bigr),$$
from which we deduce
$$\limsup_{x \to +\infty} \biggl\lvert x^{-\sigma_1}s_1 \int_1^x t^{s_1-1}\bigl(A(t) -L\bigr)\,dt\bigr\rvert \leqslant \varepsilon.$$
Since $\varepsilon > 0$ was arbitrary, that means the integral belongs to $o(x^{\sigma_1})$ too.
However, $F(x) \in o(x^{\sigma_1})$ does not suffice to directly deduce the convergence of
$$\int_1^\infty t^{-s_1-1} F(t)\,dt,$$
so we must argue a little more carefully. Since $F$ is constant on the interval $[n,n+1)$, we have
$$\int_n^{n+1} t^{-s_1-1}F(t)\,dt = F(n)\int_n^{n+1} t^{-s_1-1}\,dt =\frac{F(n)}{s_1}\biggl(\frac{1}{n^{s_1}} - \frac{1}{(n+1)^{s_1}}\biggr).$$
Then we can sum by parts to obtain
\begin{align}
\int_1^{N+1} t^{-s_1-1}F(t)\,dt &= \sum_{n = 1}^N \int_{n}^{n+1} t^{-s_1-1}F(t)\,dt \\
&= \frac{1}{s_1}\sum_{n = 1}^N F(n)\biggl(\frac{1}{n^{s_1}} - \frac{1}{(n+1)^{s_1}}\biggr)\\
&= \frac{1}{s_1}\Biggl(-\frac{F(N)}{(N+1)^{s_1}} + \sum_{n = 1}^N \frac{F(n) - F(n-1)}{n^{s_1}}\Biggr)\\
&= -\frac{F(N)}{s_1(N+1)^{s_1}} + \frac{1}{s_1}\sum_{n = 1}^N \frac{f(n)}{n^{s_1}}.
\end{align}
Since $F(x)\in o(x^{\sigma_1})$, the first term converges to $0$ as $N \to \infty$, and by assumption the series converges, so we have
$$\sum_{n = 1}^\infty \frac{f(n)}{n^{s_1}} = \lim_{N\to\infty} s_1\int_1^{N+1} t^{-s_1-1}F(t)\,dt,$$
where $N\in \mathbb{N}$. Now it suffices that $t^{-s_1-1}F(t) \to 0$ as $t\to\infty$ to deduce that
$$\lim_{R\to\infty} \int_1^R t^{-s_1-1}F(t)\,dt$$
exists, i.e. the integral
$$\int_1^\infty t^{-s_1-1} F(t)\,dt$$
converges, for
$$\biggl\lvert \int_1^R t^{-s_1-1}F(t)\,dt - \int_1^{\lfloor R\rfloor + 1} t^{-s_1-1}F(t)\,dt\biggr\rvert \leqslant \int_{\lfloor R\rfloor}^{\lfloor R\rfloor + 1} \lvert t^{-s_1-1} F(t)\rvert\,dt \leqslant \sup \{ \lvert t^{-s_1-1}F(t)\rvert : t \geqslant \lfloor R\rfloor\}.$$
Thus it follows that
$$\sum_{n = 1}^\infty \frac{f(n)}{n^{s_1}} = \lim_{N\to\infty} s_1 \int_1^{N+1} t^{-s_1-1} F(t)\,dt = s_1 \int_1^\infty t^{-s_1-1}F(t)\,dt.$$
