`sum_(n=2)^oo 1/(n(lnn)^3)` 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 a 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 1/(n(ln(n))^3)` , then `a_n =1/(n(ln(n))^3)` .

Then applying `a_n=f(x)` , we consider:`f(x) =1/(x(ln(x))^3)` .  

The graph of f(x) is:

 ` `

As shown on the graph above, the function `f(x)` is positive and decreasing on the finite interval `[2,oo)` . This implies 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^oo 1/(x(ln(x))^3)= lim_(t-gtoo)int_2^t 1/(x(ln(x))^3)dx`

To determine the indefinite integral of `int_2^t 1/(x(ln(x))^3)dx` , we may apply u-substitution by letting:

`u = ln(x)` and `du = 1/x dx` . 

The integral becomes: 

`int 1/(x(ln(x))^3)dx=int 1/(ln(x))^3 *1/x dx`

                            `=int 1/u^3 du`

Apply Law of exponent: `1/x^m = x^(-m)` .

`int 1/u^3 du=int u^(-3) du`

Apply Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`int u^(-3) du =u^(-3+1)/(-3+1)`

                    `=u^(-2)/(-2)`

                    `= - 1/(2u^2)`

Plug-in  `u=ln(x)` on `- 1/(2u^2)` , we get:

`int_2^t 1/(x(ln(x))^3)dx=- 1/(2(ln(x))^2)|_2^t`

Apply definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`- 1/(2(ln(x))^2)|_2^t=- 1/(2(ln(t))^2)-(- 1/(2(ln(2))^2))`

                       `=- 1/(2(ln(t))^2)+ 1/(2(ln(2))^2)`

Applying  `int_1^t 1/(x(ln(x))^3)dx=- 1/(2(ln(t))^2)+ 1/(2(ln(2))^2)` , we get:

`lim_(t-gtoo)int_1^t 1/(x(ln(x))^3)dx=lim_(t-gtoo)[- 1/(2(ln(t))^2)+ 1/(2(ln(2))^2)]`

                                        ` = 0+1/(2(ln(2))^2)`

                                       ` =1/(2(ln(2))^2)`

Note: `lim_(t-gtoo)1/(2(ln(2))^2)=1/(2(ln(2))^2)` and

`lim_(t-gtoo)- 1/(2(ln(t))^2)= [lim_(t-gtoo) 1]/[lim_(t-gtoo)2(ln(t))^2]`

                               ` =-1/oo`

                               ` =-0 or 0`

The`lim_(t-gtoo)int_2^t 1/(x(ln(x))^3)dx= 1/(2(ln(2))^2)`  implies that the integral converges.

Conclusion: The integral `int_2^oo 1/(x(ln(x))^3)` is convergent therefore the series `sum_(n=2)^oo 1/(n(ln(n))^3)`  must also be convergent.