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

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Answer 1 · 2016-06-17T17:28:08

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

#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)#

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