Let s n =∑k=1 to n[n/(n 2 +kn+k 2 )] and T n =∑k=0 to (n-1) [n/(n 2 +nk+k 2 )] for n=1,2,3....... Then A S n ‹ pi/3√3 C T n B Sn> pi/3 √3 D T n>pi /3√3

Let sn=∑k=1 to n[n/(n2+kn+k2)] and Tn=∑k=0 to (n-1)  [n/(n2+nk+k2)] for n=1,2,3....... Then

A  Sn‹ pi/3√3                          C  Tn

B  Sn> pi/3√3                           D  Tn>pi/3√3

Grade:Upto college level

1 Answers

Badiuddin askIITians.ismu Expert
148 Points
13 years ago

Hi vanya

You can find Sn by summation of series using definite intigral as the limit of sum

sn=∑k=1 to n[n/(n2+kn+k2)]

sn=1/n∑k=1 to n [n2/(n2+kn+k2)]

sn=1/n∑k=1 to n [1/(1+k/n+(k/n)2)]

sn =∫1/(1+x+x2)dx   limit 0 to 1

   after solving

sn =pi/3√3

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