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?
 A: You are close.
$\sum_{n=1}^{\infty}\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big)
$
$\begin{array}\\
\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big)
&=\tan{\Big(\pi n\sqrt[4]{1 + (-1)^n/n^2 + 4/n^4}} \Big)\\
&=\tan{\Big(\pi n(1 + (-1)^n/(4n^2) + O(n^{-4})} \Big)\\
&=\tan{\Big(\pi n( (-1)^n/(4n^2) + O(n^{-4})} \Big)\\
&=\tan{\Big(\pi ( (-1)^n/(4n) + O(n^{-3})} \Big)\\
&=\pi ( (-1)^n/(4n) + O(n^{-3}) \\
\end{array}
$
The sum converges
conditionally but not absolutely
since the first term
($\pi(-1)^n/(4n)$)
is an alternating series
of decreasing terms
(whose sum diverges)
and the second term
($O(1/n^3)$)
converges absolutely.
A: 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 ($\left|u_n\right|\sim \frac{1}{4n}$).
