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    Hi,

    I'm pretty sure this sequence doesn't have a limit but I don't know how to prove it (plus I may be wrong entirely)

    Xn = (n)^0.5 . (n + e^(-n))^-1 . sin(e^n)

    (apologies for the layout too, I don't know latex)

    thanks for any help!
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    I agree that it does not have a limit. The way I would approach it is by simplifying it using comparisons until you eventually have the sequence as greater than some obvious limit, and then you can say that the original sequence converges by the comparison test.
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    anyone else got any ideas?
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    I am quite confident my method is correct. The first step to see that (x_n) > \sqrt{n} \cdot \frac{1}{2n} \cdot \sin (e^n). However, suit yourself.
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    (Original post by Kolya)
    I agree that it does not have a limit. The way I would approach it is by simplifying it using comparisons until you eventually have the sequence as greater than some obvious limit, and then you can say that the original sequence converges by the comparison test.
    Your getting sequence mixed up with series dude. Ignore the sine part - watch the rest converge to zero. Then use pinching to finish the job.
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    Oops, yes I am. Apologies.
 
 
 
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Updated: November 11, 2008

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