Understanding the Divergence of Series 1/(nln(n)): A Quick Explanation

[frasifrasi]

why does the series 1/(nln(n)) diverge? I thought it converged since the limit goes to 0.

 

[rs1n]

No, the SEQUENCE

$$\left\{ \frac{1}{n\ln n} \right\}$$

converges because the limit of the terms go to 0.

However, the SERIES

$$\sum_{n=2}^\infty \frac{1}{n\ln n}$$

diverges using the integral test.

 

[rs1n]

For the series

$$\sum_{n=1}^\infty a_n$$

the condition that

$$\lim_{n\to\infty} a_n = 0$$

is necessary for convergence, however it is not sufficient. That is, satisfying the limit condition is not enough to conclude that the series converges.

 

[sutupidmath]

well use this link https://www.physicsforums.com/showthread.php?p=1487826#post1487826

i have proven it in both ways either by comparison test also by using the cauchy integral test that the above series diverges.

 

[frasifrasi]

I see, so

1/n will diverge since p <= 1 and 1/nln(n) is smaller than that, so it will converge as well--is that a correct comparison test?

 

[HallsofIvy]

No, it's not. if a_n converges and b_n< a_n then b_n converges. If a_n diverges and b_n> a_n then b_n diverges. If a_n diverges and b_n< a_n, you don't have any information as to whether b_n converges or not.
As rs1n said, use the integral test.