Question

How do you show whether `sum_(n=2)^oo 1/ln^3(n)` converges or diverges?

Answer 1

`sum_(n=2)^oo1/ln^3(n)` diverges by the Cauchy condensation test

Explanation:

The Cauchy condensation test states that if `f(n)` is nonnegative and nonincreasing real sequence, then `sum_(n=1)^oof(n)` converges if and only if the sum `sum_(n=0)^oo2^nf(2^n)` converges as well.

Let `f(n) = {(1/ln^3(n) if n>1),(f(n+1) if n=1):}`

As adding a finite value to a series does not change whether it converges or diverges, we can add `1/ln^3(2)` to the given series without changing the result. Doing so, we can apply the Cauchy condensation test:

`1/ln^3(2)+sum_(n=2)^oo1/ln^3(n) = sum_(n=1)^oof(n)` converges if and only if `sum_(n=0)^oo2^nf(2^n) = sum_(n=1)^oo2^n/ln^3(2^n)` converges.

Now, looking at the condensed sum, we have

`sum_(n=1)^oo2^n/ln^3(2^n) = sum_(n=1)^oo2^n/(nln(2))^3=sum_(n=1)^ooln^(-3)(2)2^n/n^3`

which diverges by the divergence test, as `lim_(n->oo)|2^n/n^3|=oo`.

As `sum_(n=0)^oo2^nf(2^n)` diverges, so does `sum_(n=1)^oof(n)=1/ln^3(2)+sum_(n=2)^oo1/ln^3(n)`, and thus so does the given series.

Answer 2

My first intuition is to show that `1/ln^3(n)>1/n` to claim that `sum_(n=2)^oo1/ln^3(n)` diverges through direct comparison, since `sum_(n=2)^oo1/n` diverges as the harmonic series.

Notice that `1/ln^3(n)=1/n(n/ln^3(n))`. From this we see that `1/ln^3(n)>1/n` if `n/ln^3(n)>1`.

To see if this is true as these series extend infinitely, we can take the infinite limit of `n/ln^3(n)`.

`lim_(nrarroo)n/ln^3(n)`

This is indeterminate in the form `oo/oo`, so we can apply L'Hospital's rule and tale the derivative of the numerator and denominator separately.

`=lim_(nrarroo)1/(3ln^2(n)1/n)=lim_(nrarroo)n/(3ln^2(n))`

Reapplying L'Hospitals (we can see a pattern forming):

`=lim_(nrarroo)1/(6ln(n)1/n)=lim_(nrarroo)n/(6ln(n))`

L'Hospital's once more:

`=lim_(nrarroo)1/(6 1/n)=lim_(nrarroo)n/6=oo`

Thus we've seen that `n/ln^3(n)` will be far greater than `1`, and that we can state that `1/ln^3(n)=1/n(n/ln^3(n))>1/n` for sufficiently large values of `n`.

Through direct comparison, `sum_(n=2)^oo1/ln^3(n)` diverges.