Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 15

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.