I ask this because I noticed the partial sum ∑xn=11√n is very close to 2√x, 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(n−1), 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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