# 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?

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Use the Limit Comparison Test, with $\sum 1/n^{3/2}$ (for small $x$, $\tan x\approx x$). –  David Mitra Jan 13 '14 at 0:29
sorry my question was not that "edit" changes it. tan (1/n) is in numerator –  nothingobvious Jan 13 '14 at 0:29
now it is ok.... –  nothingobvious Jan 13 '14 at 0:32
now help me to decide the convergence of the series. –  nothingobvious Jan 13 '14 at 0:38

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.

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Is the 2n limit 1??? I think it is 0 –  nothingobvious Jan 13 '14 at 0:51
@nothingobvious Woops, found a typo. Take a look at it now. –  Nick Peterson Jan 13 '14 at 0:52
how can you say that limt is 1 still please check again ...now it is $\infty$ –  nothingobvious Jan 13 '14 at 0:56
@nothingobvious Nope. It is correct now. –  Nick Peterson Jan 13 '14 at 0:57
sorry...now it is ok... –  nothingobvious Jan 13 '14 at 0:58

$\tan(x)$ 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.

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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

$\displaystyle \sum_{n=3}^{\infty} \dfrac{\tan(1/n)}{\sqrt{n}} \le \sum_{n=3}^{\infty} \dfrac{2}{n^{3/2}}$

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