Studying the convergence of a series I would like to study the convergence of the series:
$$ \sum u_{n}$$ where $$ u_{n}=a^{s_{n}}$$
$$ s_{n}=\sum_{k=1}^n \frac{1}{k^b}$$
$$ a,b<1$$
We have:
$$ u_{n}=\exp((\frac{n^{1-b}}{1-b}+o(n^{1-b}))\ln(a))=\exp(\frac{\ln(a)n^{1-b}}{1-b}+o(n^{1-b}))$$
However  $$\exp(o(n^{1-b}))$$ must be specified. So how can I determine:
 $$ o(n^{1-b})=s_{n}-\frac{n^{1-b}}{1-b}=\sum_{k=1}^n \frac{1}{k^b}-\frac{n^{1-b}}{1-b}$$ ?
 A: The $o(n^{1-b})$ term would be enough to conclude, but anyway, $t\mapsto 1/t^b$ is decreasing on $t\gt0$ and $b\lt1$ hence
$$
s_n\leqslant\sum_{k=1}^n\int_{k-1}^k\frac{\mathrm dt}{t^b}=\int_0^n\frac{\mathrm dt}{t^b}=\frac{n^{1-b}}{1-b}.
$$
Likewise,
$$
s_n\geqslant\sum_{k=1}^n\int_{k}^{k+1}\frac{\mathrm dt}{t^b}=\int_1^{n+1}\frac{\mathrm dt}{t^b}=\frac{(n+1)^{1-b}-1}{1-b}\geqslant\frac{n^{1-b}-1}{1-b}.
$$
Hence, if $a\lt1$, 
$$
u_n=\mathrm e^{s_n\log(a)}\leqslant\exp\left(\log(a)\frac{n^{1-b}-1}{1-b}\right)=C\exp(-cn^{1-b}),
$$
for some finite positive $c$ and $C$ which I will let you write down.
Edit One sees that $s_n=\dfrac{n^{1-b}}{1-b}+O(1)$ hence $u_n=\displaystyle\exp\left(\log(a)\frac{n^{1-b}}{1-b}+O(1)\right)$. Some more work yields $s_n=\dfrac{n^{1-b}}{1-b}-r_n$, where $r_n\to r$ and $r$ is finite, positive, and defined by
$$
r=\int_{0}^{+\infty}\left(\frac1{t^b}-\frac1{\lceil t\rceil^b}\right)\mathrm dt.
$$
Thus,
$$
u_n=\displaystyle\exp\left(\log(a)\frac{n^{1-b}}{1-b}-\log(a)r-v_n\right),
$$
with $(v_n)$ decreasing, $v_n\gt0$ and $v_n\to0$.
