absolute convergence Watch

DH.
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Show  \displaystyle \sum_{n=1}^\infty \dfrac{n!(2n)!\sin(n^{2013})}{(3  n)!} converges absolutely

I tried this using the ratio test, but ended up trying to figure out what the limit of  \displaystyle \left | \dfrac{\sin((n+1)^{2013})}{\sin(  n^{2013})} \right| which I have no idea :\

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if I say  a_n = \dfrac{n!(2n)!\sin(n^{2013})}{(3  n)!} and consider  |a_n| then obviously  0 < |a_n| \leq \dfrac{n!(2n)!}{(3n)!} and if I prove  \dfrac{n!(2n)!}{(3n)!} converges using the ratio test then this would be sufficient to say that  \displaystyle \sum_{n=1}^\infty a_n converges absolutely right? As this is easily possible.
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around
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Your edit has it right.
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DH.
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(Original post by around)
Your edit has it right.
thanks for the fast response
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