Sn = 3/1^3 + 5/(1^3+2^3+3^3) + 7/(1^3+2^3+3^3).... upto n terms, then lim n - infinity Sn

4

its easy

general term is T_n = 4 * (2n+1)/(n²(n+1)²)

i hope you know the formula of 1³ + 2³ + 2³ .... = n²(n+1)²/4

now split the general term as 

numerator = 2n+1 = (n+1) + (n) 

T_n = 4* [ 1/n²(n+1)   +   1/n(n+1)²]

again split the numerator

this time 1 = n+1 - n

=> T_n = 4* [ 1/n²  - 1/n(n+1)  +  1/n(n+1)   - 1/(n+1)²]

=> T_n = 4* [ 1/n²  - 1/(n+1)²]

if n-> infinity

then the subsequent terms gets cancelled and the last term tends to zero as n is in denominator

=> S_n = T1  + T2  +  T3  ......

=> S_n = 4*[1 - 1/4  + 1/4 .....]

=> S_n = 4

got it!!