`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.

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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=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`

            li`m_(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