Why Does Sum 1/(n log n) Diverge? Explained

[alice.w]

Homework Statement

Classify this sum as convergent or divergent:

E (n=2 to infinity) of 1/(n log n)

Homework Equations

The Attempt at a Solution

Using Cauchy's integral test I have integrated it to obtain:

[ log (log x)] with the boundaries of infinity and 2. When I graph this function, it clearly converges, but the answer says it's divergent! Any help explaining this?

Thank you :)

 

[Tedjn]

Are you sure that the function log(log x) is bounded from its graph? You know that it grows very slowly, because its derivative is 1/(x log x).

 

[alice.w]

Are you sure that the function log(log x) is bounded from its graph? You know that it grows very slowly, because its derivative is 1/(x log x).

That is true. So does this mean it diverges but extremely slowly?

If it converges, would it simply reach a point and stop growing? (sorry if this seems an obvious question)

 

[Tedjn]

If the series converges, which means the integral you did is finite, it means that log(log x) evaluated at infinity must be finite. This means that the function log(log x) must be bounded at infinity. When a function is always increasing, as in this case, you can picture two ways that the function is bounded at infinity. Either the function reaches a maximum value and then stays flat (i.e. stops growing) or the function keeps growing toward some finite value but never gets there. This means the function never stops growing, but just doesn't grow fast enough. If log(log x) is bounded at infinity, it must be the second case, because its derivative 1/(x log x) is never equal to 0, so the function log(log x) never stops growing.

However, it is true that log(log x) is not bounded at infinity. That means that, even though it grows extremely slowly, it will eventually grow larger than any finite number. Why is this true? It is because log x is not bounded at infinity. So, for any finite N, we can find M such that log M > N. Then, we can find x such that log x > M. So log(log x) > N.

The next question we can ask is, why is log x not bounded at infinity? The answer to that question depends on how log x is defined. One common way to define it is just as the integral of 1/x from 1 to x. In that case, we can see that there is a lower Riemann sum formed from the sum from 2 to infinity of 1/n (the harmonic series), for which there exist elementary proofs by grouping terms which show that this sum is unbounded. This proves that log x is not bounded.

 

[gomunkul51]

Evaluate the integral you got (this is the whole point of the integral test !): when x -> infinity, log(log(x)) -> infinity.
Thus, the series DIVERGES.

 

[HallsofIvy]

If it converges, would it simply reach a point and stop growing? (sorry if this seems an obvious question)

No, it doesn't. It would mean that it grows more and more slowly and never goes above some upper (or below some lower) bound.