Let s_n=∑k=1 to n[n/(n²+kn+k²)] and T_n=∑k=0 to (n-1) [n/(n²+nk+k²)] 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

Hi vanya

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

s_n=∑k=1 to n[n/(n²+kn+k²)]

s_n=1/n∑k=1 to n [n²/(n²+kn+k²)]

s_n=1/n∑k=1 to n [1/(1+k/n+(k/n)²)]

s_{n =∫1/(1+x+x²)dx   limit 0 to 1}

   after solving

s_{n =}pi/3√3

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