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Thread: Show that sin(a_n) converges

  1. #1
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    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.
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    Just notice that if $\displaystyle 0\le x$ then $\displaystyle \sin(x)\le x$.
    Use the basic comparison test.
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    Quote Originally Posted by Plato View Post
    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.
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    Quote Originally Posted by NonCommAlg View Post
    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 $\displaystyle \sin(x)<x$.
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    Quote Originally Posted by Plato View Post
    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 $\displaystyle a_n \to 0.$
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