# Thread: Show that sin(a_n) converges

1. ## Show that sin(a_n) converges

Show that if $\displaystyle \sum{a_n}$ is a convergent positive-term series, then the series $\displaystyle \sum{sin(a_n)}$ also converges.

2. Just notice that if $\displaystyle 0\le x$ then $\displaystyle \sin(x)\le x$.
Use the basic comparison test.

3. Originally Posted by Plato
Just notice that if $\displaystyle 0\le x$ then $\displaystyle \sin(x)\le x$.
Use the basic comparison test.
but $\displaystyle \sin a_n$ is not necessarily positive! we can remove this difficulty by using the inequality $\displaystyle |\sin x| \leq |x|.$

so if $\displaystyle \sum a_n$ is convergent with $\displaystyle a_n \geq 0,$ then $\displaystyle \sum \sin a_n$ is absolutely convergent and hence convergent.

4. Originally Posted by NonCommAlg
but $\displaystyle \sin a_n$ is not necessarily positive! we can remove this difficulty by using the inequality $\displaystyle |\sin x| \leq |x|.$
Actually $\displaystyle \sin(a_n)>0$ for almost all $\displaystyle n$.
It was given that $\displaystyle (a_n)$ is a positive sequence and we know that $\displaystyle (a_n)\to 0$.
Therefore, for almost all $\displaystyle n$ we have $\displaystyle 0< a_n<\frac{\pi}{2}$ which implies $\displaystyle 0<\sin(a_n)<a_n$.
So you are mistaken. Comparison works.

P.S. If $\displaystyle 0<x$ then $\displaystyle \sin(x)<x$.

5. Originally Posted by Plato
Actually $\displaystyle \sin(a_n)>0$ for almost all $\displaystyle n$.
It was given that $\displaystyle (a_n)$ is a positive sequence and we know that $\displaystyle (a_n)\to 0$.
Therefore, for almost all $\displaystyle n$ we have $\displaystyle 0< a_n<\frac{\pi}{2}$ which implies $\displaystyle 0<\sin(a_n)<a_n$.
So you are mistaken. Comparison works.

P.S. If $\displaystyle 0<x$ then $\displaystyle \sin(x)<x$.
you're right. i ignored the fact that $\displaystyle a_n \to 0.$