Sunday, 11 June 2017

sequences and series - What is limxtoinfty2sqrtxsumxn=11oversqrtn?





I ask this because I noticed the partial sum xn=11n is very close to 2x, so close in fact that it appears their difference approaches a constant value, like Hx and lnx. However, when I put this limit as is into Wolfram, it said the limit diverged.



But, I found a way to transform the limit into an infinite sum, by using the transformation f(x)=f(0)+xn=1f(n)f(n1), an application of telescoping series to partial sums.



Thus,
lim



This sum converges according to Wolfram by comparison test, and according to me by the integral test, but what does it converge to? It converges incredibly slowly; my best guess is \approx 1.458


Answer



May be, we could use generalized harmonic numbers since \sum_{n=1}^x {1\over \sqrt{n}}=H_x^{\left(\frac{1}{2}\right)} and use the asymptotics for large values of n H_x^{\left(\frac{1}{2}\right)}=2 \sqrt{x}+\zeta \left(\frac{1}{2}\right)+\frac 1 {2\sqrt x}+O\left(\frac{1}{x^{3/2}}\right)


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