Question

Using the integral test, how do you show whether `sum 1/(nln(3n))` diverges or converges from n=1 to infinity?

Answer 1

Diverges.

Explanation:

Let `f(x)=1/(xln3x)`. Now, the integral test tells us that if `f(x)` is positive, continuous, and decreasing on `[k, oo)`, the convergence or divergence of `int_k^oof(x)dx` will tell us about the convergence or divergence of `sum_(n=k)^oof(n)`.

So, we have `sum_(n=1)^oo1/(nln3n)` and we will be evaluating `int_1^oo1/(xln3x)dx`.

First, let's find the general antiderivative for `intdx/(xln3x).`

Let `u=ln(3x), du=(3/(3x))dx=dx/x`

So, we then have `int(du)/u=ln|u|=ln|ln(3x)|`

Now,

`int_1^oodx/(xln3x)=lim_(t->oo)int_1^tdx/(xln3x)=lim_(t->oo)(ln|ln3x|)|_1^t`

Evaluate the limit:

`lim_(t->oo)(ln(ln3t)-ln(ln3))=oo-ln(ln3)`

We can drop the absolute value bars. Not working with any negatives.

The integral diverges; therefore, so does the series.