<s>[INST] studying the series $\sum_\limits{n=0}^\infty \tan (\frac{n}{1+n^3})$ studying the series $\sum_\limits{n=0}^\infty \tan (\frac{n}{1+n^3})$.
The series has got a sense for each natural number.
$1+n^3>n, \forall n \in N \Rightarrow \frac{n}{1+n^3}<1,\forall n \in N\Rightarrow \tan \frac{n}{1+n^3} $ tends to 0 as $n \in N$ increases.
So all the elements are positive and tends to 0.
$$\tan \frac{n}{1+n^3} \sim \frac{\sin\frac{n}{1+n^3} }{\cos\frac{n}{1+n^3}} \sim \frac{\frac{n}{1+n^3} }{\cos\frac{n}{1+n^3}} \sim \frac{n}{1+n^3} <\frac{n}{n^3}= \frac{1}{n^2}$$
$\sum \frac{1}{n^2}$ converges $\Rightarrow $ the original series converges.
Is it right?
 [/INST] From $$x\leq\tan x\leq \frac{4x}{\pi} \quad \left(0\leq x \leq \frac{\pi}{4}\right)$$
We get $$0\leq \frac{n}{1+n^3}\leq \tan \left(\frac{n}{1+n^3}\right) \leq \frac{4n}{\pi(1+n^3)} \leq \frac{4}{\pi n^2}$$
So original series converges by comparison test.
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