Convergence of this series which gives no result in root test Is the series whose general term is $$\frac{\tan \frac{1}{n}}{\sqrt n}$$ convergent? I have tried for root test but the limit is 1 so no decision taken. How to check for convergence of this series?
 A: Remember that
$$
\tan\left(\frac{1}{n}\right)=\frac{\sin\left(\frac{1}{n}\right)}{\cos\left(\frac{1}{n}\right)},
$$
and therefore
$$
\lim_{n\to\infty}\frac{\tan\left(\frac{1}{n}\right)}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}\cdot\frac{1}{\cos\left(\frac{1}{n}\right)}=1\cdot\frac{1}{1}=1,
$$
since $\frac{1}{n}\to0$ as $n\to\infty$.
Using this, you can prove that
$$
\lim_{n\to\infty}\frac{\ \frac{\tan\left(\frac{1}{n}\right)}{\sqrt{n}}\ }{\frac{1}{n^{3/2}}}=1.
$$
In light of this, the Limit Comparison Test tells us that the two series
$$
\sum_{n=1}^{\infty}\frac{\tan\left(\frac{1}{n}\right)}{\sqrt{n}}\qquad\text{and}\qquad\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}
$$
have the same convergence behavior.
A: $\tan(x)$ is convex on $(0,\pi/2)$ and so we know that on $[0,\pi/4]$
$$
|\tan(x)|\le\frac4\pi|x|
$$
Thus,
$$
\frac{\tan\left(\frac1n\right)}{\sqrt{n}}\le\frac4\pi\frac1{n^{3/2}}
$$
and use the p-test.
A: For $n\ge 3$ (to get into the first quadrant),
$\tan(1/n) = \dfrac{\sin(2/n)}{1+\cos(2/n)} \le \sin(2/n) \le 2/n$
Hence
$\displaystyle \sum_{n=3}^{\infty} \dfrac{\tan(1/n)}{\sqrt{n}} \le  \sum_{n=3}^{\infty} \dfrac{2}{n^{3/2}}$
