Calculus  ratio test
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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 
calculus  ratio test
Posted by COFFEE on Sunday, July 29, 2007 at 6:32pm. 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) 
asked by COFFEE on July 29, 2007 
calculus  interval of convergence
infinity of the summation n=0: ((n+2)/(10^n))*((x5)^n) .. my work so far. i used the ratio test = lim (n>infinity)  [((n+3)/(10^(n+1)))*((x5)^(n+1))] / [((n+2)/(10^n))*((x5)^n)]  .. now my question is: was it ok for me to
asked by COFFEE on July 30, 2007 
calculus  interval of convergence
infinity of the summation n=0: ((n+2)/(10^n))*((x5)^n) .. my work so far. i used the ratio test = lim (n>infinity)  [((n+3)/(10^(n+1)))*((x5)^(n+1))] / [((n+2)/(10^n))*((x5)^n)]  .. now my question is: was it ok for me to
asked by COFFEE on July 29, 2007 
CALC 2
In the following series x is a real number. In each case, use the ratio test to determine the radius of convergence of the series. Analyze the behavior at the endpoints in order to determine the interval of convergence. a.
asked by Bae on April 13, 2014

CALC 2 pls help!!
In the following series x is a real number. In each case, use the ratio test to determine the radius of convergence of the series. Analyze the behavior at the endpoints in order to determine the interval of convergence. a.
asked by BAE on April 14, 2014 
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In the following series x is a real number. In each case, use the ratio test to determine the radius of convergence of the series. Analyze the behavior at the endpoints in order to determine the interval of convergence. a.
asked by BAE on April 14, 2014 
Calc Please Help
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? and lim x>0 = + infinity? lim x>1 (x^2  5x + 6)/(x^2  3x + 2) I get 2/0, so lim x> 1+ =  infinity? and lim x>1 = +
asked by Chelsea on October 14, 2010 
Calc. Limits
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? and lim x>0 = + infinity? lim x>1 (x^2  5x + 6)/(x^2  3x + 2) I get 2/0, so lim x> 1+ =  infinity? and lim x>1 = +
asked by Chelsea on October 13, 2010 
calc
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? and lim x>0 = + infinity? lim x>1 (x^2  5x + 6)/(x^2  3x + 2) I get 2/0, so lim x> 1+ =  infinity? and lim x>1 = +
asked by Chelsea on October 13, 2010