Question

How do you use the integral test to determine if `ln2/2+ln3/3+ln4/4+ln5/5+ln6/6+...` is convergent or divergent?

Answer 1

It is divergent. See explanation.

Explanation:

First write the series:

`ln2/2+ln3/3+ln4/4+...=sum_(n=2)^ooln(n)/n`

Before getting into the integral test, we must assure two things first: for the integral test to apply to `sum_(n=N)^oof(n)` we must have `a_n>0` for the given interval and `f(n)` must be decreasing on the same interval.

Both of these are true since `ln(n)>0` and `n>0` on `n in [2,oo)`, so `ln(n)/n>0` on the same interval.

Furthermore, `ln(n)` grows slower than `n` so we see that `ln(n)/n` is decreasing because `n` overpowers `ln(n)` in the numerator. You can also show this by taking the derivative of `ln(n)/n` and showing it's always negative on `n in [2,oo)`.

So, we see the integral test applies. The integral test states that if the two aforementioned conditions are met, then for `sum_(n=N)^oof(n)`, evaluate the improper integral `int_N^oof(x)dx`.

If the integral converges to a real, finite value, then the series converges. If the integral diverges, then the series does too.

So, we take the integral `int_2^ooln(x)/xdx`. Take the limit as it goes to infinity:

`int_2^ooln(x)/xdx=lim_(brarroo)int_2^bln(x)/xdx`

Letting `u=ln(x)`, so `du=1/xdx`:

`=lim_(brarroo)int_ln(2)^ln(b)ucolor(white).du`

`=lim_(brarroo)[1/2u^2]_ln(2)^ln(b)`

`=lim_(brarroo)1/2ln^2(b)-1/2ln^2(2)`

As `brarroo`, we see that `ln(b)rarroo`, so:

`=oo`

The integral diverges. Thus, we see that `sum_(n=2)^ooln(n)/n` diverges as well.