sum_(n=1)^oo ln(n)/n^2 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.

The Integral test is applicable if f is positive and decreasing function on infinite interval [k, oo) where kgt= 1 and a_n=f(x) . Then the series sum_(n=k)^oo a_n converges if and only if the improper integral int_k^oo f(x) dx converges. If the integral diverges then the series also diverges.
For the given series sum_(n=1)^oo ln(n)/n^2 , the a_n =ln(n)/n^2 .
Then applying a_n=f(x) , we consider:
f(x) =ln(x)/x^2
The graph of f(x) is:

As shown on the graph, f is positive on the finite interval [1,oo) . To verify of the function will eventually decreases on the given interval, we may consider derivative of the function.
Apply Quotient rule for derivative: d/dx(u/v) = (u'* v- v'*u)/v^2 .
Let u = ln(x) then u' = 1/x
      v = x^2 then v' = 2x
Applying the formula,we get:
f'(x) = (1/x*x^2- 2x*ln(x))/(x^2)^2
       = (x-2xln(x))/x^4
       =(1-2ln(x))/x^3
Note that 1-2ln(x) lt0 for larger values of x which means f'(x) lt0 .Based on the First derivative test, if f'(x) has a negative value then the function f(x) is decreasing for a given interval I . This confirms that the function is ultimately decreasing as x-gt oo . Therefore, we may apply the Integral test to confirm the convergence or divergence of the given series.
We may determine the convergence or divergence of the improper integral as:
int_1^oo ln(x)/x^2dx= lim_(t-gtoo)int_1^t ln(x)/x^2dx
To determine the indefinite integral of int_1^t ln(x)/x^2dx , we may apply integration by parts: int u *dv = u*v - int v* du
u = ln(x) then du = 1/x dx . 
dv = 1/x^2dx then v= int 1/x^2dx = -1/x  
Note: To determine v, apply Power rule for integration int x^n dx = x^(n+1)/(n+1).
int 1/x^2dx =int x^(-2)dx
                =x^(-2+1)/(-2+1)
                = x^(-1)/(-1)
                = -1/x
The integral becomes: 
int ln(x)/x^2dx=ln(x)(-1/x) - int (-1/x)*1/xdx
                    = -ln(x)/x - int -1/x^2dx
                    =-ln(x)/x + int 1/x^2dx
                    =-ln(x)/x + (-1/x)
                    = -ln(x)/x -1/x
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
-ln(x)/x -1/x|_1^t =[-ln(t)/t -1/t] -[-ln(1)/1-1/1]
                      =[-ln(t)/t -1/t] -[-0-1]
                      =[-ln(t)/t -1/t] -[-1]
                      = -ln(t)/t -1/t +1
Apply int_1^tln(x)/x^2dx= -ln(t)/t -1/t +1 , we get:
lim_(t-gtoo)int_1^tln(x)/x^2dx=lim_(t-gtoo) [-ln(t)/t -1/t +1]
                                  = -0 -0 +1
                                = 1
Note: lim_(t-gtoo) 1=1
         lim_(t-gtoo) 1/t = 1/oo or 0
      lim_(t-gtoo) -ln(t)/t =[lim_(t-gtoo) -ln(t)]/[lim_(t-gtoo) t]
                             =-oo/oo
Apply L' Hospitals rule:
lim_(t-gtoo) -ln(t)/t =lim_(t-gtoo) -(1/t)/1
                        =lim_(t-gtoo) -1/t
                        = -1/oo or 0
The  lim_(t-gtoo)int_1^tln(x)/x^2dx =1  implies that the integral converges.
Conclusion: The integral int_1^oo ln(x)/x^2dx   is convergent therefore the series sum_(n=1)^ooln(n)/n^2 must also be convergent.