Approximation of a summation by an integral

Solution 1:

One way to visualize what your are doing is with Riemann Sums. Assuming your function is increasing (it depends on $n$ and $\beta$) we have that $$\int_{a-1}^{b} f(x) dx \leq \sum_{k=a}^b f(k) \leq \int_{a}^{b+1} f(x) dx$$ From a more general perspective, we can apply the Euler-MacLaurin Formula, which says that $$\sum_{n=a}^b f(n) = \int_a^b f(x) dx + \frac{f(a)+f(b)}{2} + \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}(f^{(2k-1)}(b)-f^{(2k-1)}(a))+R$$ or that $$\sum_{n=a}^b f(n) \sim \int_a^b f(x) dx + \frac{f(a)+f(b)}{2} + \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}(f^{(2k-1)}(b)-f^{(2k-1)}(a))$$ Where $B_{k}$ is the $k$th Bernoulli Number (all the odd Bernoulli numbers are just zero)

The error term $R$ is very small (and depends on $a,b,$ and $f$) so we can usually ignore it unless you absolutely need it. See the linked page for an explicit formula.

Edit:
Per the OP's request, the reason we can neglect the error term is that we have the general form of the Euler-MacLaurin Formula as such: $$\sum_{n=a}^b f(n) = \int_a^b f(x) dx + \frac{f(a)+f(b)}{2} + \sum_{k=1}^{\color{red}{\left\lfloor \frac p2 \right\rfloor}} \frac{B_{2k}}{(2k)!}(f^{(2k-1)}(b)-f^{(2k-1)}(a))+R$$ In the formula I wrote above we take $p \to \infty$. We further note (see the Wikipedia page) that the remainder $R$ is bounded as such: $$|R| \leq \frac{2 \zeta(p)}{(2 \pi)^p}\int_a^b |f^{(p)}(x)|dx$$
Now, there are two parts of this. The fraction clearly goes to zero as $p \to \infty$, because the Riemann Zeta function is decreasing and bounded while the denominator grows without bounds. Thus, all we have to be able to say is that the integral doesn't blow up to $+$ or $-$ infinity as we differentiate more and more (i.e. let $p \to \infty$). Most functions you work with will satisfy this criteria, but you can check this explicitly if you so desire.

Edit 2:
As noted by @SangchulLee in the comments, for fixed $\beta$ and fixed $x$ the integral in the OP's error term appears to grow super-exponentially. While the fraction $\frac{2\zeta(p)}{(2\pi)^p}$ definitely helps decrease this growth a bit, the error term might need to be watched carefully here.

Solution 2:

If we write $s = 1/(\beta-1)$, your sum and integral can be re-written as

$$ \sum_{i=0}^{n-1} \left( \frac{n}{n-i} \right)^s = n^s \sum_{k=1}^n \frac{1}{k^s}, \qquad \int_{0}^{n-1} \left( \frac{n}{n-x} \right)^s \, dx = n^s \int_{1}^{n} \frac{dx}{x^s}. $$

In this case, the Euler-Maclaurin formula provides a way of estimating the difference within $\mathcal{O}(n^{-K})$ for any prescribed exponent $K$. To see this, let $\tilde{B}_k(x) = B_k(x - \lfloor x \rfloor)$ be the periodic Bernoulli polynomials. Then

\begin{align*} \int_{1}^{n} \frac{dx}{x^s} &= \left( \int_{[1,n]} \frac{d\lfloor x \rfloor}{x^s} + \int_{[1,n]} \frac{d\tilde{B}_1(x)}{x^s} \right) \\ &= \sum_{k=1}^{n} \frac{1}{k^s} + \int_{[1,n]} \frac{d\tilde{B}_1(x)}{x^s} \\ &= \sum_{k=1}^{n} \frac{1}{k^s} + \left[ \frac{\tilde{B}_1(x)}{x^s} \right]_{1^-}^{n} + s \int_{1}^{n} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx \\ &= \sum_{k=1}^{n} \frac{1}{k^s} - \frac{1 + n^{-s}}{2} + s \int_{1}^{n} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx \end{align*}

However, there is an issue with this form. Indeed, if we keep using $[1, n]$ as the domain of integration, the error term of the Euler-Maclaurin formula never vanishes as $n \to \infty$. This is because for each fixed $K$,

$$ \int_{1}^{n} \frac{\tilde{B}_1(x)}{x^{s+K}} \, dx = \Theta(1) \qquad \text{as } n \to \infty $$

To resolve issue, let us assume $s > 0$ and we split the last integral as the difference of two:

\begin{align*} \int_{1}^{n} \frac{dx}{x^s} &= \sum_{k=1}^{n} \frac{1}{k^s} -\frac{1 + n^{-s}}{2} + s \int_{1}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx - s \int_{n}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx \end{align*}

We note that integration by parts easily checks the asymptotics $\int_{n}^{\infty} \frac{\tilde{B}_K(x)}{x^{s+K}} \, dx = \mathcal{O}(n^{-s-K})$ for each fixed $K$. Then letting $n \to \infty$ together with the extra assumption $s > 1$, this yields

\begin{align*} s \int_{1}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx = \frac{1}{2} + \frac{1}{s-1} - \zeta(s). \end{align*}

This continues to hold for $s > 0$ by the principle of analytic continuation. Plugging this back and simplifying in terms of the sum,

\begin{align*} \sum_{k=1}^{n} \frac{1}{k^s} &= \int_{1}^{n} \frac{dx}{x^s} + \frac{1 + n^{-s}}{2} + s \left( \zeta(s) - \frac{1}{2} - \frac{1}{s-1} \right) + s \int_{n}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx. \end{align*}

Then we can continue the procedure to extract more terms: for each fixed $K$,

\begin{align*} \sum_{k=1}^{n} \frac{1}{k^s} &= \int_{1}^{n} \frac{dx}{x^s} + \frac{1 + n^{-s}}{2} + s \left( \zeta(s) - \frac{1}{2} - \frac{1}{s-1} \right)\\ &\qquad - \sum_{j=2}^{K} \frac{\Gamma(s+j-1)}{\Gamma(s)} \frac{B_j}{j!} \frac{1}{n^{s+j-1}} + \underbrace{\frac{\Gamma(s+K)}{K!\Gamma(s)} \int_{n}^{\infty} \frac{\tilde{B}_K(x)}{x^{s+K}} \, dx}_{\mathcal{O}(n^{-s-K-1})} \end{align*}

as $n \to \infty$.