#1+n (log_e n)^2 > n (log_e n)^2#
also
#1/(1+n (log_e n)^2 ) < 1/( n (log_e n)^2)#
So if #sum_{i=2}^infty 1/( n (log_e n)^2)# is convergent then
# sum_{i=2}^infty 1/( 1+n (log_e n)^2)# will be convergent
but #int_2^n dx/( x (log_e x)^2) ge sum_{i=3}^{n+1} 1/( n (log_e n)^2)#
because #1/( x (log_e x)^2)# is monotonically decreasing
and #int dx/( x (log_e x)^2) = -1/log_e(x)#
also
#{
(lim_{x->oo}-1/log_e(x) = 0),
(lim_{n->oo}1/( n (log_e n)^2)=0)
:}#
So
# sum_{i=2}^infty 1/( 1+n (log_e n)^2)# is convergent
Comparison between
#int_2^n dx/( x (log_e x)^2)# and #sum_{i=3}^{n+1} 1/( n (log_e n)^2)#
