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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

To find the convergence of the series `sum_(n=2)^oo (ln(n))/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=k)^oo a_n` converges if and only if the improper integral `int_k^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 (ln(n))/n^p ` , we have:

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

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

The `f(x)` satisfies the conditions for integral test when `pgt0` . We set-up the improper integral as:

`int_2^oo (ln(x))/x^pdx`

Apply integration by parts: `int u dv = uv - int v du.`

Let: `u=ln(x)` then `du = 1/xdx`

       `dv = 1/x^p dx`

Then , `v = int dv`

              `=int 1/x^p dx `

              `= int x^(-p) dx`

             `= x^(-p+1)/(-p+1)`

The indefinite integral will be:

`int (ln(x))/x^pdx = ln(x)x^(-p+1)/(-p+1)- intx^(-p+1)/(-p+1) *1/x dx`

                    `= ln(x)x^(-p+1)/(-p+1)-1/(-p+1) int (x^(-p)x)/x dx`

                   `= ln(x)x^(-p+1)/(-p+1)-1/(-p+1) intx^(-p) dx `        

                  `= ln(x)x^(-p+1)/(-p+1)-1/(-p+1) *x^(-p+1)/(-p+1)`

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




The definite integral will only be finite if `1-p<0 or pgt1` .

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

Approved by eNotes Editorial Team