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

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

Grade:12

1 Answers

Shivam Dimri
43 Points
8 years ago

its easy

general term is Tn = 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) 

Tn = 4* [ 1/n2(n+1)   +   1/n(n+1)2]

again split the numerator

this time 1 = n+1 - n

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

=> Tn = 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

=> Sn = T1  + T2  +  T3  ......

=> Sn = 4*[1 - 1/4  + 1/4 .....]

=> Sn = 4

got it!!

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