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

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`


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


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


                 `=(ln(x)x^(-p+1)(-p+1)-x^(-p+1))/(-p+1)^2|_2^oo`


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` .

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