# Evaluating the limit of a sum using integration

One of the first results we learn in definite integral is that if $$f(x)$$ is Riemann integrable in $$(0,1)$$ then we have $$\lim_{n \to \infty}\dfrac{1}{n}\sum_{i=1}^{n}f\Big(\dfrac{i}{n}\Big) = \int_{0}^{1}f(x)dx$$.

I was playing around with this to see if this can be generalized and I found the following. We can rewrite the above result as

$$\lim_{n \to \infty}\frac{1}{1+1+\ldots\text{n-times}}\sum_{i=1}^{n}1\times f\Big(\frac{1+1+\ldots\text{i-times}}{1+1+\ldots\text{n-times}}\Big) = \int_{0}^{1}f(x)dx.$$

The LHS can be written in the general form given below and we ask ourselves for which sequence $$a_i$$ does the following hold

$$\lim_{n \to \infty}\frac{1}{a_1 + a_2 + \ldots + a_n}\sum_{i=1}^{n}a_i f\Big(\frac{a_1 + a_2 + \ldots + a_i}{a_1 + a_2 + \ldots + a_n}\Big) =\int_{0}^{1}f(x)dx.$$

Trivially this holds for $$a_i = c$$ where $$c$$ is a non-zero constant and the above result is the case when $$c=1$$. I also observed that this holds for sequence of natural numbers $$a_i = i$$ since

$$\lim_{n \to \infty}\frac{2}{n^2+n}\sum_{i=1}^{n}i f\Big(\frac{i^2+i}{n^2+n}\Big) =\int_{0}^{1}f(x)dx.$$

Experimentally, this also holds for the sequence of prime numbers $$a_i = p_n$$ and also for the sequence of composite numbers $$c_n$$.

Question: What are the necessary and sufficient conditions on $$a_i$$ for the above relation to hold?

Related question

Here is a clumsy criterion:

Proposition. Let $$(a_n)$$ be a sequence of positive numbers and write $$s_n = \sum_{i=1}^{n} a_i$$ for the partial sums. Then the followings are equivalent:

1. For any Riemann-integrable $$f : [0, 1] \to \mathbb{R}$$, $$\lim_{n\to\infty} \sum_{i=1}^{n} f\left(\frac{s_i}{s_n}\right)\frac{a_i}{s_n} = \int_{0}^{1}f(x) \, \mathrm{d}x.$$

2. $$\max\{a_1,\cdots,a_n\}/s_n \to 0$$ as $$n\to\infty$$.

This statement is kind of dumb, since $$\max\{a_1,\cdots,a_n\}/s_n$$ represents the length of the largest subinterval of the partition in OP's scheme. Then (2) simply requires that the partition becomes finer as $$n$$ grows.

Proof. Write $$\|\Pi\|$$ for the mesh-size of the partition $$\Pi$$. If $$f : [0, 1] \to \mathbb{R}$$ is Riemann-integrable and $$\Pi_n$$ is a sequence of partitions of $$[0, 1]$$ with $$\|\Pi_n\|\to 0$$, then the associated Riemann sum converges to the integral $$\int_{0}^{1} f(x)\,\mathrm{d}x$$ as $$n\to\infty$$.

• $$(2)\Rightarrow(1)$$ : If we choose $$\Pi_n = \{s_i/s_n\}_{i=0}^{n}$$, then $$\|\Pi_n\| = \max\{a_1,\cdots,a_n\}/s_n$$, and so, (1) follows.

• $$(1)\Rightarrow(2)$$ : We prove the contrapositive. Assume that (2) does not hold. Then we can find an interval $$[a, b] \subseteq [0, 1]$$ with $$a < b$$ and a subsequence $$(n_k)$$ such that $$[a, b]$$ is always contained in one of the subintervals of $$\Pi_{n_k}$$.

Indeed, negating (2) tells that $$\limsup_{n\to\infty} \|\Pi_n\| > 0$$, thus by passing to a subsequence, we can assume that $$\|\Pi_j\| \geq \epsilon > 0$$ holds for all $$j$$, for some $$\epsilon > 0$$. Next, for each $$j$$, pick a subinterval $$I_j$$ of $$\Pi_j$$ having length $$> \epsilon$$. Then we may appeal to the compactness of $$[0, 1]$$ to extract a further subsequence $$\{\Pi_k\}$$ for which $$\bigcap_k I_k$$ is an interval of positive length. (For instance, pick a further subsequence such that the left-endpoints of $$I_k$$'s converge.)

Once such $$[a, b]$$ and $$\Pi_{n_k}$$ are chosen, simply pick $$f$$ as a Riemann-integrable function which is supported on $$(a, b)$$ and $$\int_{0}^{1} f(x) \, \mathrm{d}x \neq 0$$. Then from $$\sum_{i=1}^{n_k} f(s_i/s_{n_k}) (a_i/s_{n_k}) = 0$$, we know that this Riemann sum does not converge to the integral of $$f$$.

• That's a rather simple criterion. – Nilotpal Kanti Sinha May 9 at 6:09