an = 2n * Sin[1/(n+1)]
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lim an (as n tends to infinity)
= lim 2n * sin[1/(n+1)]
= lim sin[1/(n+1)] / [1/(2n)] <--indeterminate form 0/0, so use L'Hopital's Rule.
= lim [cos[1/(n + 1)] * (-1)/(n + 1)^2] / [(-1)/(2n^2)]
= lim cos[1/(n + 1)] * 2n^2/(n + 1)^2
= cos(0) * 2
= 2, which is finite. So the sequence converges.
= lim 2n * sin[1/(n+1)]
= lim sin[1/(n+1)] / [1/(2n)] <--indeterminate form 0/0, so use L'Hopital's Rule.
= lim [cos[1/(n + 1)] * (-1)/(n + 1)^2] / [(-1)/(2n^2)]
= lim cos[1/(n + 1)] * 2n^2/(n + 1)^2
= cos(0) * 2
= 2, which is finite. So the sequence converges.