Question

Convergence
the series from n=1 to inf of 1/n^2 *tan(1/n)

I know the series converges, I just need to prove it. I think I use the limit comparison test, but no matter what I compare it to, I keep getting a limit of 0, which doesn't work with the limit comparison test. Help please.

Answer 1

\(\tan(\frac{1}{n})\) goes to zero, so there may be some trouble here
if you are sure it converges, it looks like it might be a set up for the integral test

Answer 2

that's what I was afraid of. only because I haven't used it in forever.
how do I know if an integral converges?
I know the rule is
If \[\int\limits_{1}^{\inf}f(x) dx\] converges, then \[\sum_{}^{}a_n\] converges
where \[a_n=\frac{ \tan(\frac{ 1 }{ n }) }{ n^2 }\]

Answer 3

hold on
is the tangent in the numerator???

Answer 4

yeah, in my book it looks like
\[\sum_{1}^{\inf}\frac{ 1 }{ n^2 }\tan(\frac{ 1 }{ n })\]

Answer 5

if that is the case you have nothing to worry about
i thought it was in the denominator
since \(\tan(\frac{1}{n})\to 0\) as \(n\to \infty\) you can say that the numerator is, from some point on, less than one
so the whole thing is less that \(\frac{1}{n^2}\) and that some converges for sure
so you can use the comparison test

Answer 6

*that SUM converges

Answer 7

i.e.
\[\sum\frac{1}{n^2}\tan(\frac{1}{n})<\sum\frac{1}{n^2}\]

Answer 8

are you saying limit comparison test or basic comparison test? I guess I understand why the basic comparison test works because 1/n^2 converges since it's a p series.
but when I use the limit comparison test:
\[\lim_{n \rightarrow \infty} \frac{ \tan \frac{ 1 }{ n } }{ n^2 }\times n^2=0\] right?

Answer 9

no the comparison is to \(\frac{1}{n^2}\)
i am not sure where you got the other \(n^2\) from

Answer 10

\(\tan(\frac{1}{n})<1\) so and so \(\frac{\tan(\frac{1}{n})}{n^2}<\frac{1}{n^2}\)

Answer 11

at least for n larger than 2 or 3

Answer 12

sorry I got the other n^2 from
\[\lim_{n \rightarrow \infty}\ \frac{ \frac{ \tan \frac{ 1 }{ n } }{ n^2 } }{ \frac{ 1 }{ n^2 }}\] which is what the limit comparison is right?
\[\lim_{n \rightarrow \infty}\frac{ a_n }{ b_n }\]