Absolute and conditional convergence of a series with $\tan$ $$\sum_{n=1}^{\infty}\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big)$$
I tried to do something like this but in the end it don't help:
$$\sum_{n=1}^{\infty}\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big) = 
 \sum_{n=1}^{\infty}\tan{\Big(\pi n  \sqrt[4]{1 + \Big(\frac{(-1)^n}{n^2} + \frac{4}{n^4}}\Big)} = \\
\sum_{n=1}^{\infty}\tan{\Big(\pi n \Big(1 + \frac{(-1)^n n^2 + 4}{4n^4} \Big) \Big)}.$$
Could you give me tips how to continue solution or solve in another way?

First, don't try to sum the series when you are transforming its terms. Work on the terms, find equivalents or developments, and then, apply a theorem stating the status of the series.
Here, you took the right start :
$$\sqrt[4]{n^4+(-1)^nn^2+4} = n\left(1+\frac{(-1)^n}{n^2}+\frac{4}{n^4}\right)^{\frac{1}{4}} = n\left(1+\frac{(-1)^n}{4n^2}+O\left(\frac{1}{n^4}\right)\right)$$
so
\begin{align}\tan\left(\pi \sqrt[4]{n^4+(-1)^nn^2+4}\right) &= \tan\left(n\pi+\frac{(-1)^n}{4n}+O\left(\frac{1}{n^3}\right)\right) \\ &= \tan\left(\frac{(-1)^n}{4n}+O\left(\frac{1}{n^3}\right)\right) \\ &= \frac{(-1)^n}{4n}+O\left(\frac{1}{n^3}\right)\end{align}
Therefore $u_n$ is the sum of two terms, one, $\frac{(-1)^n}{4n}$, leads to a convergent, but not absolutely convergent, series, the other, $O\left(\frac{1}{n^3}\right)$, is the term of an absolutely convergent series.
Conclusion : $\sum u_n$ is convergent, but not absolutely convergent ($\left|u_n\right|\sim \frac{1}{4n}$).
