Tuesday, January 10, 2017

sequences and series - Find the value of : limlimitsnrightarrowinftyleft(2sqrtnsumlimitsnk=1frac1sqrtkright)



How to find limn(2nnk=11k) ?



And generally does the limit of the integral of f(x) minus the sum of f(x) exist?
How to prove that and find the limit?


Answer



Use n=nk=1(kk1), then

2nnk=11k=nk=1(2k2k11k)=nk=11k(kk1)2=nk=11k((kk1)(k+k1)(k+k1))2=nk=11k(k+k1)2



This shows the limit does exist and limn(2nnk=11k)=k=11k(k+k1)2.




The value of this sums equals ζ(12)1.4603545. This value is found by other means, though:
2nnk=11k=2n(ζ(12)ζ(12,n+1))ζ(12)12n+o(1n)


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