`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.

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marizi eNotes educator| Certified Educator

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

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