1+n (log_e n)^2 > n (log_e n)^21+n(logen)2>n(logen)2
also
1/(1+n (log_e n)^2 ) < 1/( n (log_e n)^2)11+n(logen)2<1n(logen)2
So if sum_{i=2}^infty 1/( n (log_e n)^2)∞∑i=21n(logen)2 is convergent then
sum_{i=2}^infty 1/( 1+n (log_e n)^2)∞∑i=211+n(logen)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)∫n2dxx(logex)2≥n+1∑i=31n(logen)2
because 1/( x (log_e x)^2)1x(logex)2 is monotonically decreasing
and int dx/( x (log_e x)^2) = -1/log_e(x)∫dxx(logex)2=−1loge(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)
![enter image source here]()
