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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

7 years ago


Answers : (1)


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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7 years ago

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