Let f(x)=1/(xln3x)f(x)=1xln3x. Now, the integral test tells us that if f(x)f(x) is positive, continuous, and decreasing on [k, oo)[k,∞), the convergence or divergence of int_k^oof(x)dx∫∞kf(x)dx will tell us about the convergence or divergence of sum_(n=k)^oof(n)∞∑n=kf(n).
So, we have sum_(n=1)^oo1/(nln3n)∞∑n=11nln3n and we will be evaluating int_1^oo1/(xln3x)dx∫∞11xln3xdx.
First, let's find the general antiderivative for intdx/(xln3x).∫dxxln3x.
Let u=ln(3x), du=(3/(3x))dx=dx/xu=ln(3x),du=(33x)dx=dxx
So, we then have int(du)/u=ln|u|=ln|ln(3x)|∫duu=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.
