ln2/2+ln3/3+ln4/4+ln5/5+ln6/6+... Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

Integral test is applicable if f is positive, continuous and decreasing function on infinite interval [k,oo) where k>=1 and f(n)=a_n . Then the series sum_(n=k)^ooa_n converges or diverges if and only if the improper integral int_k^oof(x)dx converges or diverges.
Given series is ln2/2+ln3/3+ln4/4+ln5/5+ln6/6+........
The series can be written as sum_(n=1)^ooln(n+1)/(n+1)
Consider f(x)=ln(x+1)/(x+1)
Refer the attached graph for f(x),
From the graph, we observe that the function is positive, continuous and decreasing for x>=2
We can also determine whether f(x) is decreasing by finding the derivative f'(x), such that f'(x)<0 for x>1
Since the function satisfies the conditions for the integral test , we can apply the same.
Now let's determine the convergence or divergence of the integral int_1^ooln(x+1)/(x+1)dx
int_1^ooln(x+1)/(x+1)dx=lim_(b->oo)int_1^bln(x+1)/(x+1)dx
Let's first evaluate the indefinite integral,intln(x+1)/(x+1)dx
Apply integral substitution:u=ln(x+1)
=>du=1/(x+1)dx
intln(x+1)/(x+1)dx=intudu
=u^2/2+C , where C is a constant
Substitute back u=ln(x+1)
=1/2[ln(x+1)]^2+C
lim_(b->oo)int_1^ooln(x+1)/(x+1)dx=lim_(b->oo)[1/2(ln(x+1))^2]_1^oo
=lim_(b->oo)[1/2(ln(b+1))^2]-[1/2(ln(2))^2]
lim_(x->oo)(x+1)=oo
lim_(u->oo)ln(u)=oo
=1/2oo^2-1/2(ln(2))^2
=oo-1/2(ln(2))^2
=oo
Since the integral int_1^ooln(x+1)/(x+1)dx diverges, so the given series also diverges as per the integral test.