Convergence of $\sum\frac{\alpha_n}{n^z}$ for $\Re(z)>s$ when $\sum\frac{\alpha_n}{n^s}$ converges The problem is

Show that $\sum^\infty_{n=1}\dfrac{\alpha_n}{n^z}$, $\Re(z)>s$ converges when  $\sum^\infty_{n=1}\dfrac{\alpha_n}{n^s}$ converges. ($z,\alpha_n\in\mathbb C$, $s\in\mathbb R$)

I think this problem is an elementary application of Riemann zeta function, but I merely can prove this with a stronger condition : Absolute Convergence of $\sum^\infty_{n=1}\frac{\alpha_n}{n^s}$.
Maybe I should use the inequality below : 
$$|n^z|=n^{\Re(z)}>n^s$$
Thanks.
 A: If $z$ is real, since $\sum \frac{\alpha_n}{n^s}$ is bounded, and $\frac{1}{n^{z-s}}$ is decreasing to 0, then by Dirichlet test,  $\sum \frac{\alpha_n}{n^z}$ converges. The proof would end here. However, the possibility that $z$ can be complex complicate things a little bit. We remedy this below.

Denote $z-s = \sigma + it$, with $\sigma > 0$. Let $${S_n} = \sum\limits_{k = 1}^n {\frac{{{\alpha _k}}}{{{k^s}}}} $$ Since it converges, it is bounded.
Summation by part gives
$$\sum\limits_{k = 1}^n {\frac{{{\alpha _k}}}{{{k^z}}}}  = \sum\limits_{k = 1}^n  {\frac{{\frac{{{\alpha _k}}}{{{k^s}}}}}{{{k^{z - s}}}}}  = \frac{{{S_n}}}{{{n^{z - s}}}} - \sum\limits_{k = 1}^{n - 1} {{S_k}  \left[ {\frac{1}{{{{(k + 1)}^{\sigma + it}}}} - \frac{1}{{{k^{\sigma + it}}}}} \right]} $$
Since $\Re(z-s) > 0$, the term outside the summation tends to 0. We only need to prove that the series
$$\sum\limits_{k = 1}^{\infty} \left| {{{(\frac{1}{{k + 1}})}^{\sigma  + it}} - {{(\frac{1}{k})}^{\sigma  + it}}} \right| $$ converges

Denote $A_k = \left| {{x^{\sigma  + it}} - {y^{\sigma  + it}}} \right| - \left| {{x^\sigma } - {y^\sigma }} \right|$, where $x=1/(k+1), y=1/k$.
Then
$$\begin{aligned}|A_k| &=\sqrt {{x^{2\sigma }} + {y^{2\sigma }} - 2{x^\sigma }{y^\sigma }\cos (t\ln x - t\ln y)}  - \sqrt {{x^{2\sigma }} + {y^{2\sigma }} - 2{x^\sigma }{y^\sigma }} \\
& = \frac{{2{y^\sigma }\left[ {1 - \cos (t\ln x - t\ln y)} \right]}}{{\sqrt {1 + {{(\frac{y}{x})}^{2\sigma }} - 2{{(\frac{y}{x})}^\sigma }\cos (t\ln x - t\ln y)}  + \sqrt {1 + {{(\frac{y}{x})}^{2\sigma }} - 2{{(\frac{y}{x})}^\sigma }} }} \\
&\le {C_1}k{y^\sigma }\left[ {1 - \cos (t\ln \frac{x}{y})} \right] \le \frac{{{C_2}}}{{{k^{1 + \sigma }}}}\\
\end{aligned}$$
where we have used $1-\cos x = O(x^2)$ as $x\to 0$. The $k$ comes from the denominator.
Since $\sigma>0$, $\sum A_k$ is absolutely convergent. Note that
$$\sum\limits_{k = 1}^n {\left| {{{(\frac{1}{{k + 1}})}^\sigma } - {{(\frac{1}{k})}^\sigma }} \right|}  = 1 - {(\frac{1}{{n + 1}})^\sigma }$$ so the series $$\sum\limits_{k = 1}^{\infty} {\left| {{{(\frac{1}{{k + 1}})}^\sigma } - {{(\frac{1}{k})}^\sigma }} \right|} $$ converges.
Therefore by the convergence of $\sum A_k$, we have the desired convergence of $$\sum\limits_{k = 1}^{\infty} \left| {{{(\frac{1}{{k + 1}})}^{\sigma  + it}} - {{(\frac{1}{k})}^{\sigma  + it}}} \right| $$ This completes the proof.
