Student Question

`sum_(n=2)^oo 1/(n(lnn)^p)` Find the positive values of p for which the series converges.

Expert Answers

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To find the convergence of the series `sum_(n=2)^oo 1/(n(ln(n))^p) ` where `pgt0` (positive values of `p` ), we may apply integral test.

Integral test is applicable if f is positive, continuous, and decreasing function on an interval and let `a_n=f(x)` . Then the infinite series` sum_(n=1)^oo a_n` converges if and only if the improper integral `int_1^oo f(x) dx ` converges to a real number. If the integral diverges then the series also diverges.

For the infinte series series `sum_(n=2)^oo 1/(n(ln(n))^p)` , we have:

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

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

The `f(x)` satisfies the conditions for integral test based on the following reasons:

-` f(x)` is continuous since` x(ln(x))^p !=0` for any x-value on the interval `[2,oo)`

-` f(x)` is positive since `1/(x(ln(x))^p)gt0 ` for any x-value on the interval `[2,oo).`

-` f(x)` is decreasing since `f'(x)` is negative  for large value of `x` . It eventually decreases at the tail of the series.

To evaluate the convergence of the series using integral test, we set-up the improper integral as: 

`int_2^oo 1/(x(ln(x))^p)dx`

Apply u-substitution by letting: `u=ln(x)` `u=ln(x)` then `du = 1/xdx` , `a=ln(2)` and `b=oo` .

`int_(ln(2))^oo 1/(x(ln(x))^p) dx=int_(ln(2))^oo 1/(ln(x))^p *1/xdx`

                           `= int_(ln(2))^oo 1/u^pdu`

                          ` =int_(ln(2))^oo u^-p dx`

                          `= u^(-p+1)/(-p+1)|_(ln(2))^oo `  



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


                               `= 1/oo-1/(-p+1)1/(ln(2))^(p-1)`

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

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

The integral converges to a real number when `pgt1` .

Thus, the series `sum_(n=2)^oo 1/(n(ln(n))^p)` converges whenever `pgt1` .

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