`sum_(n=3)^oo 1/(nlnn[ln(lnn)]^p)` Find the positive values of p for which the series converges.

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To find the positive values of p in which the series `sum_(n=3)^oo 1/(nln(n)(ln(ln(n)))^p)` , we may apply the integral test.

Integral test is applicable if f is positive, continuous, and decreasing function and `a_n=f(x)` . The infinite series `sum_(n=k)^oo a_n` converges if and only of the improper integral `int _k^oo...

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To find the positive values of p in which the series `sum_(n=3)^oo 1/(nln(n)(ln(ln(n)))^p)` , we may apply the integral test.

Integral test is applicable if f is positive, continuous, and decreasing function and `a_n=f(x)` . The infinite series `sum_(n=k)^oo a_n` converges if and only of the improper integral `int _k^oo f(x)dx ` converges to a finite number. If the integral diverges then the series also diverges.

For the infinite series `sum_(n=3)^oo 1/(nln(n)(ln(ln(x)))^p)` , we have:

`a_n=1/(nln(n)(ln(ln(n)))^p)`

Then, `f(x) =1/(xln(x)(ln(ln(x)))^p)`

The `f(x)` satisfies the conditions for integral test for the interval `[3,oo)`

We set-up the improper integral as:

`int_3^oo1/(xln(x)(ln(ln(x)))^p) dx.`

Apply u-substitution by letting `u = ln(x)` then `du =1/xdx` .

`int 1/(xln(x)(ln(ln(x)))^p) dx=int 1/(ln(x)(ln(ln(x)))^p)* 1/xdx`

                      `=int 1/(u(ln(u))^p) du`

Apply another set of substitution: let` v =ln(u)` and `dv = 1/u du` .

`int 1/(u(ln(u))^p) du=int 1/(ln(u))^p* 1/u du`

                       `=int 1/v^p* dv`

                       ` =int v^(-p) dv`

                        `= v^(-p+1)/(-p+1)`

Recall `u =ln(x) ` and `v = ln(u)`  then `v =ln(ln(x))` .

`v^(-p+1)/(-p+1)=(ln(ln(x)))^(-p+1)/(-p+1)|3^oo`

The integral is finite when `-p+1lt0` or `pgt1` .

Note: When` (ln(ln(x)))` has positive power on the numerator side then the integral diverges. 

 When` (ln(ln(x)))` has negative power on the numerator side then the integral converges. 

Thus, the series `sum_(n=3)^oo 1/(nln(n)(ln(ln(n)))^p)` converges when `pgt1` .

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