We know that $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$, $\sum_{i=1}^{n}\frac{1}{i}=\psi_{0} (n+1)-\psi_{0} (1)$, where $\psi_{0}(x)$ is the digamma function.

My problem is,

(1).Is there a transformation such that it maps $x \to \frac{x(x+1)}{2}$ and $\frac{1}{x} \to \psi_{0}(x+1)$, and map a smooth $f(x)$ into another smooth function $g(x)$, such that $g(x)-g(x-1)=f(x)$ ? When I mention transformation, I mean an operator or algorithm for me to get $g(x)$ from $f(x)$.

(2). Surely $g(x)$ if exists, it is not unique because $g(x)+C$ also satisfy the condition. Let's take $g(x)+C$ and $g(x)$ as the same case. Is there another smooth $h(x)\not = g(x)+C$ satisfying this condition?

The problem came when I tried to evaluate $\sum_{i=1}^{n} \sqrt{i}$, I'd like to represent it by integral form. Thanks for attention!


You can use the zeta function and the Hurwitz zeta function to write your sum. As a more general case of your summation, we can have the following representation

$$\sum_{i=1}^{n} i^s = \sum_{i=1}^{\infty}i^s - \sum_{i=0}^{\infty}(i+n+1)^s = \zeta(-s) - \zeta(-s, n+1)\,.$$ Now, substituting $s= -\frac{1}{2}$ in the above identity yields

$$\sum_{1}^{n} i^{\frac{1}{2}} = \zeta\left(-\frac{1}{2}\right) - \zeta\left(-\frac{1}{2}, n+1\right) \,.$$

See convergence issues of the zeta function and the Hurwitz zeta function.

  • $\begingroup$ Thank you for your reply! I care more about the first two problems :is there any useful operator to perform the inverse difference? $\endgroup$ – Golbez Aug 24 '12 at 13:15

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