Question

Help with absolute convergence of series (-1)^n tan(1/n) needed.

Answer 1

\[\sum_{n=1}^{Infinity} (-1)^n \tan \frac{ 1 }{ n }\]

Answer 2

$$
\sum_{n=1}^\infty \left|(-1)^n\tan\frac1n\right|=\sum_{n=1}^\infty\tan\frac1n
$$since \(0<\frac1n\le1\) and \(\tan\cdot\) is positive on this interval

Answer 3

now consider $$\tan\frac1n\approx\frac1n$$for larger \(n\) so you should use a limit comparison test

Answer 4

Oh, I see now. Thank you :)

Answer 5

$$\lim_{n\to\infty}\frac{\tan\frac1n}{\frac1n}=\lim_{n\to\infty}1=1$$therefore \(\sum\limits_{n=1}^\infty\tan\frac1n\) diverges by the limit comparison test and our series is not absolutely convergent