Question

Does `sum_{n=2} 1 / (1 + n ( Ln(n) )^2)` converges or diverges from n=2 to infinity?

Answer 1

`sum_{i=2}^infty 1/( 1+n (log_e n)^2)` is convergent

Explanation:

`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)`