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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
B Sn> pi/3√3 D Tn>pi/3√3
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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