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Does the series sigma evaluated from n equals 1 to infinity of 1 over n times the quantity the natural log of n squared converge or diverge?
Diverge!
Does the series n factorial times 1 divided by e to the n converge or diverge?
Use the ratio test: Let an = n!/enlim n→∞ |an+1/an|= lim n→∞ |[(n + 1)!/en+1]/(n!/en)= lim n→∞ |[(n + 1)!/en+1](en/n!)= lim n→∞ |[(n + 1)n!en]/(enn!e)= (1/e) lim n→∞ (n + 1) = ∞, so the given series diverges.
What is 6 and 1 over 2 plus 1 and 1 over 2 equals 4 x n?
8 = 4n n = 2
What is n over 18 x 6 over n to the 4th power?
(n/(18*6))/n4=(n/108)/n4 ;Multiply 18 and 6(n/108)*(1/n4) ;Multiply by the reciprocal of n4, which is just 1 over n4n/(n4*108) ;The n4 will go in the bottom of the fraction1/(108n3) ;n over n4 will give you 1 over n3
What is the value of n when 6 over n times 5 over n-1 equals 1 over 3 showing all work?
If you mean: 6/n times 5/n-1 = 1/3 Then: 30/n2-n = 1/3 Multiplying both sides by n2-n: 30 = n2-n/3 Multiplying both sides by 3: 90 = n2-n Subtracting 90 from both sides: 0 = n2-n-90 or n2-n-90 = 0 Solving the above quadratic equation: n = -9 or n =10 If n is of a material value its more likely to be 10 Note that n2 means n squared
