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No. ∑(1/n) diverges. It is the special infinite series known as the "harmonic series."

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Q: Does 1 over n converge
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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 -1n converge?

No, -1^n does not converge as it alternates between -1 and 1 for different values of n. This oscillation prevents the sequence from approaching a specific limit.


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(n&pi;) &pi; = picos([2n+1]&pi;/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&rarr;&infin; |an+1/an|= lim n&rarr;&infin; |[(n + 1)!/en+1]/(n!/en)= lim n&rarr;&infin; |[(n + 1)!/en+1](en/n!)= lim n&rarr;&infin; |[(n + 1)n!en]/(enn!e)= (1/e) lim n&rarr;&infin; (n + 1) = &infin;, 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&gt;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&gt;max{N,L}+1, |n-L|=n-L&gt;1&gt;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