Doubts on asymptotic criterion for $\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}n^{a}\tan^{-1}\bigg(\frac{1}{n^a}\bigg)-e^{1/n}$ with $a>0$. I have to valuate the character of the following series:
$$\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}n^{a}\tan^{-1}\bigg(\frac{1}{n^a}\bigg)-e^{1/n}$$
with $a>0$.
I have thought that definitely the sequence $a_n$ is made by constant signed terms. So I can apply the asymptotic criterion to study the series.
$=n^{a}\tan^{-1}\big(\frac{1}{n^a}\big)-e^{1/n}\\
=n^{a}\big(\frac{1}{n^a}-\frac{1}{3n^{3a}}+o(\frac{1}{n^{3a+1}})\big)-1-\frac{1}{n}+o(\frac{1}{n})\\
=1-\frac{1}{3n^{2a}}+o(\frac{1}{n^{2a+1}})-1-\frac{1}{n}+o(\frac{1}{n})\\
=\frac{1}{3n^{2a}}+o(\frac{1}{n^{2a+1}})-\frac{1}{n}+o(\frac{1}{n})\\
\color{red}{=\frac{1}{3n^{2a}}-\frac{1}{n}+o(\frac{1}{n})}$
The red passage is right? 
I have thought that since $2a+1>1$ then I can put the $o(\frac{1}{n^{2a+1}})$ into  $o(\frac{1}{n})$.
Now:
if $a\leq 1/2$ then I have $\frac{1}{3n^{2a}}-\frac{1}{n}+o(\frac{1}{n})=\frac{1}{3n^{2a}}+o(\frac{1}{n})$, so the corresponding series diverges and also the original one.
if $a>1/2$ then  $\frac{1}{3n^{2a}}-\frac{1}{n}+o(\frac{1}{n})=-\frac{1}{n}+o(\frac{1}{n})$ and again the series diverges.
My overall attempt is right?
Edit : I understand there are some problems in the final part with the little $o$. Can someone help me
 A: First note that
$$
\arctan x = x - \frac{{x^3 }}{3} + \mathcal{O}(x^5 )\quad \text{ and }\quad e^x  = 1 + x + \mathcal{O}(x^2 )
$$
as $x\to 0$. Thus,
\begin{align*}
a_n  &= n^a \left( {\frac{1}{{n^a }} - \frac{1}{{3n^{3a} }} + \mathcal{O}\!\left( {\frac{1}{{n^{5a} }}} \right)} \right) - \left( {1 + \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) \\ & =  - \frac{1}{{3n^{2a} }} - \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^{4a} }}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)
\end{align*}
as $n\to +\infty$. Now if $a = \frac{1}{2} + \varepsilon  > \frac{1}{2}$, then it is readily seen that
$$
a_n  =  - \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^{1 + 2\varepsilon } }}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) \sim - \frac{1}{n}
$$
as $n\to +\infty$, whence the series diverges by the limit comparison test (using the fact that the harmonic series diverges). If $
a = \frac{1}{2}$, we find
$$
a_n  =  - \frac{4}{{3n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) \sim - \frac{4}{{3n}}
$$
as $n\to +\infty$, whence the series diverges by the limit comparison test. If $
0 < a < \frac{1}{2}$, then the first term will dominate, i.e.,
$$
a_n  \sim  - \frac{1}{{3n^{2a} }}
$$
as $n\to +\infty$, and the series diverges by the limit comparison test and the $p$-series test.
