calculus

determine whether the series converges of diverges

the sum from k=1 to infinity of

sin(e^-k)

I'm not sure where to start..

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asked by sarah
  1. As k becomes large, e^-k becomes much less than 1, and sin(e^-k) approaches e^-k
    The sum of the series 1 + 1/e + 1/e^2 converges to
    1 /(1 - 1/e)= 1.582
    High-order terms of the series
    sin(e^-k) will behave similarly, but the sum of the entire series will be somethat less than 1.582.

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    posted by drwls

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