Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 14

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.