9
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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 $f$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.