Ttdfjt

One of the first results we learn in definite integral is that if $f(x)$ is Riemann integrable in $(0,1)$ then we have $lim_n to inftydfrac1nsum_i=1^nfBig(dfracinBig) = int_0^1f(x)dx$.

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

$$
lim_n to inftyfrac11+1+ldotstext$n$-timessum_i=1^n1times fBig(frac1+1+ldotstext$i$-times1+1+ldotstext$n$-timesBig) = int_0^1f(x)dx.
$$

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

$$
lim_n to inftyfrac1a_1 + a_2 + ldots + a_nsum_i=1^na_i fBig(fraca_1 + a_2 + ldots + a_ia_1 + a_2 + ldots + a_nBig) =int_0^1f(x)dx.
$$

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

$$
lim_n to inftyfrac2n^2+nsum_i=1^ni fBig(fraci^2+in^2+nBig) =int_0^1f(x)dx.
$$

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 $f : [0, 1] to mathbbR$,
   $$ lim_ntoinfty sum_i=1^n fleft(fracs_is_nright)fraca_is_n = int_0^1f(x) , mathrmdx. $$




2. $maxa_1,cdots,a_n/s_n to 0$ as $ntoinfty$.

This statement is kind of dumb, since $maxa_1,cdots,a_n/s_n$ 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 $f : [0, 1] to mathbbR$ 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 $int_0^1 f(x),mathrmdx$ as $ntoinfty$.

• $(2)Rightarrow(1)$ : If we choose $Pi_n = s_i/s_n_i=0^n$, then $|Pi_n| = maxa_1,cdots,a_n/s_n$, 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 $limsup_ntoinfty |Pi_n| > 0$, 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 $int_0^1 f(x) , mathrmdx neq 0$. 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 $f$.