How do you find the positive values of p for which #Sigma lnn/n^p# from #[2,oo)# converges?

Question

Christopher Allers · Accepted Answer

The series:

#sum_(n=2)^oo lnn/n^p#

is convergent for # p > 1#.

A necessary condition for a series to converge is that its general term is infinitesimal, that is: (1) #lim_(n->oo) lnn/n^p= 0# Consider the function #f(x) = lnx/x^p# and the limit: (2) #lim_(x->oo) lnx/x^p# Clearly if such limit exists it must coincide with (1). Now, for #p=0# the limit (2) is #oo#, while for #p!=0# it is in the indeterminate form #oo/oo# so we can solve it using l'Hospital's rule: # lim_(x->oo) lnx/x^p = lim_(x->oo) (d/(dx) lnx)/(d/(dx) x^p) = lim_(x->oo) (1/x)(1/(px^(p-1))) = lim_(x->oo) 1/(px^p) = { (0 " for " p > 0),(oo " for " p < 0):}# So we know that for #p<=0# the series is not convergent. To have a sufficient condition we can apply the integral test, using as test function: #f(x) = lnx/x^p#. For #p>0# such function is positive, decreasing and, as we have just seen, infinitesimal, so it satisfies the hypotheses of the integral test theorem, and the series is proven to be convergent if the improper integral: #int_2^oo lnx/x^p# also converges. Solving the indefinite integral by parts: #int lnx/x^pdx = int lnx d(x^(1-p)/(1-p)) = lnx x^(1-p)/(1-p) -int 1/x x^(1-p)/(1-p)dx = 1/(1-p)lnx/x^(p-1) -int 1/x x^(1-p)/(1-p)dx# #int 1/x x^(1-p)/(1-p)dx =1/(p-1) int x^(-p)dx = 1/(1-p)^2x^(1-p) +C = 1/(1-p)^2 1/x^(p-1) +C# So that: #int_2^oo lnx/x^p = [1/(1-p)lnx/x^(p-1) - 1/(1-p)^2 1/x^(p-1)]_2^oo# which is convergent for #p-1 > 0 => p>1# In conclusion the series is convergent for #p > 1#

Aurora Alvarado · Answer

undefined To find the positive values of $ p $ for which the series $\sum_{n=2}^{\infty} \frac{\ln n}{n^p} $ converges, we use the integral test. We integrate the function $ \frac{\ln x}{x^p} $ from 2 to infinity and check for convergence.

Let $ u = \ln x $. Then, $ du = \frac{1}{x} dx $. Substituting these into the integral:

$$ \int_{2}^{\infty} \frac{\ln x}{x^p} dx = \int_{\ln 2}^{\infty} u \cdot u^{1-p} du = \int_{\ln 2}^{\infty} u^{2-p} du $$

This integral converges if $ 2 - p > 1 $ (since it's a power function). So, $ p < 1 $.

Therefore, the positive values of $ p $ for which the series converges are $ 0 < p < 1 $.