Question

help with integral test..what does the series converge to?

Answer 1

\[\sum_{n=1}^{\infty} \frac{ \ln(n) }{ n }\]

Answer 2

\[\int\limits_{1}^{\infty} \frac{ \ln(n) }{ n} = \lim_{b \rightarrow \infty} \frac{ \ln^2(n) }{ n }\]

Answer 3

from 1 to infinity

Answer 4

is that correct so far?

Answer 5

Why the squared?

Answer 6

Drop the squared and looks ok

Answer 7

oh sorry i meant \[\lim_{b \rightarrow \infty} \frac{ \ln^2(n) }{ 2} \]

Answer 8

from 1 to infinity

Answer 9

All you should do is evaluate the limit of the interval from 1 to infinity of your function first off. Idk how your getting 2 and squares but it doesn't look right to me.

Answer 10

but for the integral test i thought i had to find the integral of ln(n)/n

Answer 11

\[\Large \int \frac{\ln(x)}{x}dx = \int \ln(x)*\frac{1}{x}dx\]
let u = ln(x), so du = 1/x*dx
\[\Large \int \frac{\ln(x)}{x}dx = \int u du\]
\[\Large \int \frac{\ln(x)}{x}dx = \frac{1}{2}u^2+C\]
\[\Large \int \frac{\ln(x)}{x}dx = \frac{1}{2}\left[\ln(x)\right]^2+C\]

Answer 12

so am i right or wrong?

Answer 13

Alright, I see why you'd think that. But the integral test is only true if your function is continuous.

Answer 14

So it would be better to use the comparison test. Ask what is this series like as far as sorts of series are concerned?

Answer 15

Are you familiar with the comparison and limit comparison tests?

Answer 16

you are right, but as @daniel.ohearn1 said, lots easier to say that the terms are larger than the well know divergent harmonic series

Answer 17

The terms must be smaller than another series that is convergent if we want to show that it is convergent. How about the square root of n divided by n? List some terms.. Know what that would be?

Answer 18

im a bit confused..

Answer 19

Is the square root of n less or more than the ln(n) what do you think?

Answer 20

more than

Answer 21

Let A(n) be your series ln(n)/n from n=1 to n=infinity

Let B(n) be another series only defined as being positive and decreasing i.e the derivative is negative if B(n)=f(x)

The comparison test says that if B(n) is convergent and A(n)<B(n), A(n) is also convergent. Make sense?

Answer 22

In other words A(n)= B(n)+ C

Answer 23

B(n) can be any convergent series, as long as it's terms are decreasing in value and all of it's terms are positive.

Answer 24

Let A(n) = ln(n)/n
Let B(n) = 1/n

summing 1/n from 1 to infinity is known to diverge. It's the harmonic series

since B(n) < A(n), when n > 2.72, and B diverges, this means A must diverge as well

wolfram comes to the same conclusion

http://www.wolframalpha.com/input/?i=sum(ln(n)%2Fn,+n+%3D+1+..+infinity)

Answer 25

@daniel.ohearn1 ok that makes sense. but then how do i find what it converges to?

Answer 26

Does look like it diverges. I thought compare it to sqrt(n)/n which is equivalent to the p-series 1/(n^1/2) and that diverges, suggesting possible divergence.. The test says that when A(n) is more than B(n) and B(n) is divergent, A(n) is also divergent. So 1/n seems a reasonable comparison. What is the range of functions B(n) could include?

Answer 27

I/n and ln(n)/n have the same first terms making them good comparable, positive, decreasing series.

Answer 28

@jim_thompson5910

Answer 29

The series diverges and I posted why that is above

Answer 30

B(n) can be 1/Cn+D where C is any integer and D is of any lesser value than C, my thought.

Answer 31

ok thank you. @daniel.ohearn1 @jim_thompson5910

Answer 32

That's true for D as n approaches infinity.