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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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