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No. ∑(1/n) diverges. It is the special infinite series known as the "harmonic series."
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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
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!
Give examples of a convergent series.?
These are some series (not the summation of series) that converge: 1/n1/n2(a/b)n if a/b < 1 or = 1sin(1/n)cos(1/n)sin(nπ) π = picos([2n+1]π/2)e-n(n+2)/n
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.
How do you prove that the sum of a convergent sequence divided by n will converge?
You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.
Does the series 1 divided by ln x converge?
Compare a series to a known series. So take the harmonic series {1/1 + 1/2 + 1/3 + ... + 1/n}, which diverges.For each number n [n>1], LN(n) < n, so 1/(LN(n)) > 1/n. So since each term in 1/LN(n) is greater than each term in the divergent series {1/n}, then the series 1/LN(n) diverges.
How do you find the harmonic and geometric mean for grouped data?
The geometric-harmonic mean of grouped data can be formed as a sequence defined as g(n+1) = square root(g(n)*h(n)) and h(n+1) = (2/((1/g(n)) + (1/h(n)))). Essentially, this means both sequences will converge to the mean, which is the geometric harmonic mean.
Prove that the sequence n does not converge?
If the sequence (n) converges to a limit L then, by definition, for any eps>0 there exists a number N such |n-L|N. However if eps=0.5 then whatever value of N we chose we find that whenever n>max{N,L}+1, |n-L|=n-L>1>eps. Proving the first statement false by contradiction.
How do you solve 1 over n SIN x?
sinx=n/1 (1)sinx=n/1(1) sin(-n)x=n(-n) six=6
What does n over 6 minus one over 6 equals one half?
n _ 1 6 6 = (n-1) 6
What is n- 1 over 5 equals 3 over 5?
N=4.
Solve for the missing number 1 and a half equals n over 3 and 1 over 6?
you are saying 1.5 is equal to n over and 1.5 is equal to 1 over 6 first of all 1.5 is not euqal to one over six. but...if 1.5 = (n/3) then n=4.5
What is the answer to 4 divided by 8 over n?
0.5
