Testing convergence of $ \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt[n]{3}-1}{\cot(\frac{1}{n})+(-1)^n} $ I'm having trouble testing the convergence of the following series:
$$
\sum_{n=1}^{\infty} (-1)^n \frac{\sqrt[n]{3}-1}{\cot(\frac{1}{n})+(-1)^n}
$$ Any advice?
 A: Let $a_n$ denote the term of the series. By the series expansion for $e^x$, we have
$$
3^{1/n} - 1 = \left(1 + \log(3) n^{-1} + O(n^{-2})\right)-1 = \log(3) n^{-1}+O(n^{-2});
$$for what it's worth, the constants are all positive. Note that for $n\ge 2$, $1/n <\pi/4$, so $\cot(1/n)>1$. Also, for $M>1$, $M+(-1)^n \ge M-1$, so $(M+(-1)^n)^{-1} \le (M-1)^{-1}$. So, we have
$$
|a_n| = \frac{3^{1/n} -1 }{\cot(1/n)+(-1)^n} \le \frac{\log(3) n^{-1}}{\cot(n^{-1})+(-1)^n} \le \frac{\log(3) n^{-1}}{\cot(n^{-1})-1} 
$$
Multiply by the conjugate:
$$
 = \frac{\log(3) n^{-1} (\cot(n^{-1})+1)}{\cot^2(n^{-1})-1}= \frac{\log(3) n^{-1} (\cot(n^{-1})+1)}{\csc^2(n^{-1})}
$$
$$
 = \frac{\log(3) \sin(n^{-1})}{n} \cdot\sin(n^{-1})\cdot\left(\cot(n^{-1})+1\right)
$$Simplify and bound:
$$
 =  \frac{\log(3) \sin(n^{-1})}{n} \cdot \left(\sin(n^{-1})+\cos(n^{-1})\right)
$$
$$
\le 2 \log(3) \frac{\sin(n^{-1})}{n} \le 2 \log(3) n^{-2};
$$here we have used $\sin(\theta)< \theta$. In conclusion, the series converges by absolute convergence and then direct comparison to something like $K n^{-2}$.
