sum_(n=2)^oo ln(n)/n^3 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=2)^oo ln(n)/n^3 ,  a_n =ln(n)/n^3 .
Then applying a_n=f(x) , we consider:
f(x) =ln(x)/x^3
The graph of f(x) is:

As shown on the graph, f is positive on the infinite interval [1,oo) . To verify of the function will eventually decreases on the given interval, we may consider the derivative of the function.
Apply Quotient rule for the derivative: d/dx(u/v) = (u'* v- v'*u)/v^2 .
Let u = ln(x) then u' = 1/x
      v = x^3 then v' = 3x^2
Applying the formula,we get:
f'(x) = (1/x*x^3- 3x^2*ln(x))/(x^3)^2
           = (x^2-3x^2ln(x))/x^6
           =(1-3ln(x))/x^4
Note that 1-3ln(x) lt0 for larger values of x which means f'(x) lt0 .Based on the first derivative test, if f'(x) lt0 then f(x) is decreasing for a given interval I. This confirms that the function is ultimately decreasing as x-> 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_2^ooln(x)/x^3dx= lim_(t-gtoo)int_2^tln(x)/x^3dx
To determine the indefinite integral of int_2^tln(x)/x^3dx , we may apply integration by parts: int u dv = uv - int v du
u = ln(x) then du = 1/x dx.
dv = 1/x^3dx then v= int 1/x^3dx = -1/(2x^2)
Note: To determine v , apply Power rule for integration int x^n dx = x^(n+1)/(n+1)
int 1/x^3dx =int x^(-3)dx
                =x^(-3+1)/(-3+1)
                  = x^(-2)/(-2)
                = -1/(2x^2)
The integral becomes: 
int ln(x)/x^3dx=ln(x) (-1/(2x^2)) - int -1/(2x^2)*1/xdx
                    = -ln(x)/(2x^2) - int -1/(2x^3)dx
                    =-ln(x)/(2x^2) + 1/2 int 1/x^3dx
                    =-ln(x)/(2x^2) + 1/2*(-1/(2x^2))
                    = -ln(x)/(2x^2) -1/(4x^2)
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
-ln(x)/(2x^2) -1/(4x^2)|_2^t=[-ln(t)/(2t^2) -1/(4t^2)] -[-ln(2)/(2*2^2) -1/(4*2^2)]
                           = [-ln(t)/(2t^2) -1/(4t^2)]-[-ln(2)/8 -1/16]
                           =-ln(t)/(2t^2) -1/(4t^2) + ln(2)/8 + 1/16
                          =-ln(t)/(2t^2) -1/(4t^2) +1/16 (ln(4) +1)
Note:ln(2)/8 + 1/16 = 1/16 (2ln(2) +1)
                          =1/16 (ln(2^2) +1)
                          =1/16 (ln(4) +1)
 Apply  int_2^tln(x)/x^3dx=-ln(t)/(2t^2) -1/(4t^2) +1/16 (ln(4) +1) , we get:
lim_(t-gtoo)int_2^tln(x)/x^3dx=lim_(t-gtoo) [ -ln(t)/(2t^2) -1/(4t^2) + 1/16(ln(4)+1)]
                                  = -0 -0+1/16(ln(4)+1)
                                  =1/16(ln(4)+1)
  Note: lim_(t-gtoo) 1/16(ln(4)+1)=1/16(ln(4)+1)
           lim_(t-gtoo) 1/(4t^2)= 1/oo or 0
            lim_(t-gtoo) ln(t)/(2t^2)=[lim_(t-gtoo) -ln(t)]/[lim_(t-gtoo) 2t^2]=-oo/oo
Apply L' Hospitals rule:
lim_(t-gtoo) ln(t)/(2t^2) =lim_(t-gtoo) (1/t)/(4t)
                     =lim_(t-gtoo) 1/(4t^2)
                     = 1/oo or 0
The  lim_(t-gtoo)int_2^tln(x)/x^3dx=1/16 (ln(4) +1) implies that the integral converges.
Conclusion: The integral int_2^ooln(x)/x^3dx   is convergent therefore the series sum_(n=2)^ooln(n)/n^3 must also be convergent.