1
$\begingroup$

I tried to prove the convergence of this series with the ratio and root test but neither worked. I also couldn't find another series to compare. I understand that $\tan\frac{1}{n}$ is decreasing and the partial sums too.

I have no idea what to try now

$$\sum _{n=1}^{\infty }(-1)^n n\tan\frac{1}{n}.$$

$\endgroup$
1
  • 2
    $\begingroup$ Hint: near $0$, $\tan(x)$ is very much like $x$ (look at the Taylor series), so your sum is essentially $\sum (-1)^n\cdot 1$, which does not converge... $\endgroup$ – paul garrett Mar 31 at 16:19
5
$\begingroup$

$$n\tan\left(\dfrac1n\right)=\frac1{\cos\left(\dfrac1n\right)}\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}$$ shows that the general term does not tend to zero.

$\endgroup$
4
  • $\begingroup$ When I take the $\underset{n\to \infty }{\text{lim}}$ the right side became $\left( \frac{1}{1}\right) \frac{0}{0}$, because $\frac{0}{0}$ I apply L'Hôpital then it goes to 1. Is it right? $\endgroup$ – Carlos Eduardo Mar 31 at 16:46
  • 1
    $\begingroup$ @CarlosEduardo Are you familiar with the classical limit $$ \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1? $$ $\endgroup$ – Gary Mar 31 at 16:53
  • 1
    $\begingroup$ @CarlosEduardo that is one way of doing it, yes. For another way (using Maclaurin series of $\sin(x)$), see my answer. $\endgroup$ – Adam Rubinson Mar 31 at 16:53
  • $\begingroup$ @Gary yes. I didn't notice I could use the it. $\endgroup$ – Carlos Eduardo Mar 31 at 17:01
1
$\begingroup$

Expanding on Yves Daoust's terse answer,

$\displaystyle\lim_{n\to\infty}\cos\left(\dfrac1n\right) = 1\neq 0,\ $ so $\ \displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\right) = \frac{1}{1} = 1,\ $ and here, we see that:

$$\displaystyle\lim_{n\to\infty}\left(\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = \displaystyle\lim_{n\to\infty}\left(\frac{\left(\dfrac1n\right)-\frac{\left(\dfrac1n\right)^3}{3} + \frac{\left(\dfrac1n\right)^5}{5}+\ldots}{\dfrac 1n}\right) = \displaystyle\lim_{n\to\infty}\left(1-\frac{\left(\dfrac1n\right)^2}{3} + \frac{\left(\dfrac1n\right)^4}{5}+\ldots\right) = 1.$$

Therefore,

$$\displaystyle\lim_{n\to\infty}\left(n\tan\left(\dfrac1n\right)\right)=\displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = \displaystyle\lim_{n\to\infty}\left(\frac1{\cos\left(\dfrac1n\right)}\right)\ \cdot\ \displaystyle\lim_{n\to\infty}\left(\frac{\sin\left(\dfrac1n\right)}{\dfrac 1n}\right) = 1\cdot 1 = 1.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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