does the infinite series defined by the natural log of n devided by n converge?

Answer 1

log(n)/n > 1/n if n>2, and the sum of 1/n diverges (this is the so-called harmonic series), so log(n)/n also diverges.

Answer 2

Alright it diverges, as many people have already said. In fact some of the posts sort of hint this as being rather trivial, and it is, but only if you know the following two points. I asssume you didn't think it trivial, or you wouldn't have wasted your precious points and thus:

1) the series 1/n diverges because of the integral test. Notice the integral of 1/n from 1 to infinity < than the sum of each term 1/n from 1 to infinity (this can be seen by graphing the curve vs. graphing a block 1/1 from 1 to 2, 1/2 from 2 to 3 etc. , but the integral of 1/n is ln(n) so it amounts to ln(infinity) - ln(1) which is infinity -0, and it diverges, and so the sum 1/n also diverges.

2) Even if you cannot immediately see that 1/n < ln(n), for n > 2. Nitice the inequality only depends on the upper half of the fractions, and futhermore that 1 is constant while ln(n) increases with out bounds, so there must be some point where the curves cross. We don't even need to find this point, just to know that at some point we are adding bigger amounts than 1/n.

So the series diverges.

PS: The limit of ln(n)/n at infinity is 0, which can be demonstrated with L'hopital's rule. which states it is equal to the limit of (1/n)/1. The series still diverges, by points 1 and 2.

Answer 3

Do you mean the terms of the series or the sum of the series? The terms converge to 0. The sum does not converge: ln(n)/n > 1/n for n > 2, so since the sum 1/3 + 1/4 + 1/5... does not converge, ln(3)/3 + ln(4)/4 + ... also does not.

Answer 4

Use the Integral Test

let f(x) = (ln x) / x
therefore f'(x) = (1 - ln x) / x^2; f'(x) < 0 when ln x > 1 (or x > e)

Since f(x) is continuous, positive and decreasing on the interval [1,infinity), then technically you can use the Integral Test.
(Albeit the function is only decreasing when x > e, this does not affect things...either way it will be only one of two things convergent or divergent)

Integral[1,infinity) { (ln x)/ x dx}

Do u substitution, u = ln x, du = 1/x
Gives...
(ln x)^2 / 2 evaluated from 1 to infinity

Take the lim t -->infinity (ln x)^2 / 2 evaluated from 1 to t

(ln t)^2 / 2 goes to infinity, hence the Integral is divergent

Since the integral is divergent, it follows that the series is also divergent.

Answer 5

Yes it converges.


Limits as n approaches infinity and in this case, k is constant.

lim ((ln n) / n) = lim ((1/n) / k(n^(k-1))) = lim (1 / k(n^k)) = 0