it is

   lim { (3-2) (4-2) (5-2) ..(3²+3*2+2²)+ ...} / { (3+2)(4+2) (5+2) ...(3²-3*2+2²)+...}

 

(n -> infinity)       

 

lim {1*2*3*4 * (n²+3)(n²+2n+4)}/ {(n-1)n(n+1)(n+2) ..7*12

(see first four terms in numerator will survive and last four in denominator will survive)

so

lim 2/7 * n⁴(1+3/n²) (1+2/n+4/n²) / n⁴(1-1/n)(1+1/n)(1+2/n) *n³

i think the denominator u typed an extra n³

but anyway if the n³ is there i think the answer is 0

otherwise the answer is 2/7