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Calculus
Determine whether the series is convergent or divergent. If it is convergent, find its sum. If it is divergent, enter NONE. sum from 1 to infinity of 1/e^2n. It is convergent, but I do not know how to solve for the sum.
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determine if the series is absolutely convergent and convergent. the sum from n=0 to infinity of ((1)^n*e^n)/(n!) I used the ratio test and said that it was absolutely convergent and convergent. is this true?
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determine if the series is absolutely convergent and convergent. the sum from n=0 to infinity of ((1)^n*e^n)/(n!) I used the ratio test and said that it was absolutely convergent and convergent. is this true?
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Determine whether the series is convergent or divergent. If it is convergent, find its sum. If it is divergent, enter NONE. sum from 1 to infinity of (5^n+4^n)/20^n It is convergent, but I do not know how to find the sum.
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asked by sarah on February 21, 2008 
Calculus  ratio test
infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n>infinity)  [(e^n+1)/((n+1)!)] / [(e^n)/(n!)]  = lim (n>infinity)  [(e^n+1)/((n+1)!)] * [(n!)/(e^n)]  = lim (n>infinity) 
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calculus
for each series determine if the series is absolutely convergent and convergent the sum from 0 to infinity of (1)^n/(the square root of (n+1)) I did the ratio test and got 1, which is less than 0 making it absolutely convergent,
asked by sarah on March 2, 2008 
calculus  ratio test
infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n>infinity)  [(e^n+1)/((n+1)!)] / [(e^n)/(n!)]  = lim (n>infinity)  [(e^n+1)/((n+1)!)] * [(n!)/(e^n)]  = lim (n>infinity) 
asked by COFFEE on July 30, 2007