Consider the function:
#f(x) = lnx/sqrtx# with #x > 0#
evaluate the derivative using the quotient rule:
#(df)/dx = ( sqrtx d/dx lnx - lnx d/dx sqrtx)/(sqrtx)^2#
#(df)/dx = ( sqrtx/x - lnx/(2sqrtx))/x#
#(df)/dx = 1/x( 1/sqrtx - lnx/(2sqrtx))#
#(df)/dx = ( 2 - lnx)/(2xsqrtx)#
For #x > e^2# we have that #lnx > 2# and then #(df)/dx < 0#, so that #f(x)# is monotone decreasing.
It follows that for #n > 7#:
#(1) " " ln(n+1)/sqrt(n+1) < lnn/sqrtn#
The limit:
#lim_(x->oo) lnx/sqrtx#
is in the indeterminate form #oo/oo# so we can solve it using l'Hospital's rule:
#lim_(x->oo) lnx/sqrtx = lim_(x->oo) (d/dx lnx)/(d/dx sqrtx) = lim_(x->oo ) (1/x) / (1/(2sqrtx)) = lim_(x->oo) 2/sqrtx = 0#
In particular for #x = n#:
#(2) " " lim_(n->oo) lnn/sqrtn = 0#
As for #n>1# we have that #lnn/sqrtn > 0# the series:
#sum_(n=1)^oo (-1)^n lnn/sqrtn#
is an alternating series, and using Leibniz' criteria, we can see that #(1)# and #(2)# prove that the series is convergent.
