Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

3 Answers 3

up vote 2 down vote accepted

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.

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

share|improve this answer

For $n\ge 3$ (to get into the first quadrant),

$\tan(1/n) = \dfrac{\sin(2/n)}{1+\cos(2/n)} \le \sin(2/n) \le 2/n$


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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.