Determine the convergence or divergence of the series. (E = sigma) infinity E (-1)^n+1 (2) / e^n - e^-n n=1
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i assume it is .. notice that it's a sequence of hyperbolic cosecant.. and from n=1 to infinity, that is monotone and bounded..
Originally Posted by element Determine the convergence or divergence of the series. (E = sigma) infinity E (-1)^n+1 (2) / e^n - e^-n n=1 Lets put this in math notation: Well for large , is strictly decreasing and goes to zero, and so the alternating series test applies and your series is convergent. RonL
yeah, i forgot to mention that
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