This question was inspired by the one at http://answers.yahoo.com/question/index;_ylt=ApfyiAkLxa3nBMvRv5dBST3sy6IX?qid=20070816104652AAunj7P
For x >=0, let f(x) = lim [1/1^x ....+1/n^x - Integral (1 to oo) dt/t^x]. Then f(x) = lim [1/1^x ....+1/n^x - (n^(1-x) -1)/(1 -x)], if x<>1, and f(1) = lim [1/1 ....+1/n - ln(n)] , so that f(1) is the Euler/Mascheroni constant. Since the function u --> 1/u^x is strictly decreasing for x>0, the limit of the sequence always exist and f is well defined. For x >1, 1- x <0, both the series and the integral converge and we have f(x) =lim [1/1^x ....+1/n^x - 1/(x -1)] = Z(x) - 1/(x -1), where Z is the Riemann Zeta function. We know Z is analytic on the complex plane, so has derivatives of all orders on (1, oo). Therefore, , f has derivatives of all orders on this set and f'(x) = Z'(x) + 1/(x-1)^2. But I couldn't conclude if f is differentiable, or at least continuous, on [0,1].
2007-08-20
07:59:41
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1 answers
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asked by
Steiner
7