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.

Images:

Image (1 of 1)