Show that sin(a_n) converges

**alexmahone** · September 7th 2010, 03:49 AM

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

**Plato** · September 7th 2010, 04:13 AM

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

**NonCommAlg** · September 7th 2010, 10:39 AM

Originally Posted by Plato

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

but $\sin a_n$ is not necessarily positive! we can remove this difficulty by using the inequality $|\sin x| \leq |x|.$

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

**Plato** · September 7th 2010, 10:54 AM

Originally Posted by NonCommAlg

but $\sin a_n$ is not necessarily positive! we can remove this difficulty by using the inequality $|\sin x| \leq |x|.$

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

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

**NonCommAlg** · September 7th 2010, 11:16 AM

Originally Posted by Plato

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

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

you're right. i ignored the fact that $a_n \to 0.$

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