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

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- Sep 7th 2010, 03:49 AMalexmahoneShow 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.

- Sep 7th 2010, 04:13 AMPlato
Just notice that if $\displaystyle 0\le x$ then $\displaystyle \sin(x)\le x$.

Use the basic comparison test. - Sep 7th 2010, 10:39 AMNonCommAlg
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. - Sep 7th 2010, 10:54 AMPlato
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 $\displaystyle \sin(x)<x$. - Sep 7th 2010, 11:16 AMNonCommAlg