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I was pondering how to solve a limit of the kind $$\lim_{n \to ∞} n^k \left(\dfrac{1}{n}\sum_{k=0}^{n-1} f \left(\dfrac{k}{n} \right) -\int_0^1 f(x)dx \right)$$

where k is chosen such that the order of infinity is just right for the limit to be non-zero but not infinite. For instance, if $f(x)=x^2$, then the limit becomes;

$$\lim_{n \to ∞} n^k \left(\dfrac{(n+1)(2n+1)}{6n^2}-\dfrac{1}{3} \right)$$

Here, k should be 1 and the final answer will be the ratio of the coefficients of the linear term in the numerator and denominator i.e. $1/2$. But, this was a relatively simple integral and I'm searching for a general solution or algorithm to calculate this kind of limit. Here, I'm assuming $f(x)$ is a well-behaved function i.e. it is a $C^∞$ function with a Taylor series which converges for $0<x<1$.

I thought of perhaps applying l'hopital's rule to the sum but the issue is that the number of terms itself is dependent of the variable with respect to which I wish to differentiate.

Then I thought of perhaps writing the Taylor series and integrating term by term, combining each term of the integration with the corresponding sum of that power in the sum. After simplifying everything, I was only left with the remainder term of the Taylor expansion on which I was unable to apply the limit properly.

How can I approach this problem and also, is it possible to decide beforehand what the value of k should be for such a limit to exist?

Cognoscenti
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1 Answers1

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if $f\in C^2[0,1]$,we have the conclusion:

$\begin{aligned}\lim_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{k=0}^nf\left(\frac{k}{n}\right)-\int_0^1f(x)\mathrm{d}x\right)=\frac{1}{2}[f(1)-f(0)]\end{aligned}$

Consider Taylor expansion of $f$

$\begin{aligned}f(x)=f\left(\frac{k}{n}\right)+f'\left(\frac{k}{n}\right)\left(x-\frac{k}{n}\right)+\frac{f''(\xi_{n,k})}{2}\left(x-\frac{k}{n}\right)^2\end{aligned}$

Let $M=\max_{x\in[0,1]}\{|f''(x)|\}$, we can find that

$\begin{aligned}&n\left|\frac{1}{n}\sum_{k=0}^nf\left(\frac{k}{n}\right)-\int_0^1f(x)\mathrm{d}x+\sum_{k=0}^n\int_{\frac{k-1}{n}}^\frac{k}{n}f'\left(\frac{k}{n}\right)\left(x-\frac{k}{n}\right)\mathrm{d}x\right|\\ &=n\left|\sum_{k=0}^n\int _{\frac{k-1}{n}}^\frac{k}{n}f\left(\frac{k}{n}\right)\mathrm{d}x-\sum_{k=0}^n\int_\frac{k-1}{n}^\frac{k}{n}f(x)\mathrm{d}x+\sum_{k=0}^n\int_{\frac{k-1}{n}}^\frac{k}{n}f'\left(\frac{k}{n}\right)\left(x-\frac{k}{n}\right)\mathrm{d}x\right|\\ &=n\left|\sum_{k=0}^n\int_\frac{k-1}{n}^\frac{k}{n}\left[f\left(\frac{k}{n}\right)-f(x)-f'\left(\frac{k}{n}\right)\left(x-\frac{k}{n}\right)\right]\mathrm{d}x\right|\\ &=\frac{n}{2}\left|\sum_{k=0}^n\int_\frac{k-1}{n}^\frac{k}{n}f''(\xi_{n,k})\left(x-\frac{k}{n}\right)^2\mathrm{d}x\right|\\ &\leq\frac{Mn}{2}\left|\sum_{k=0}^n\int_\frac{k-1}{n}^\frac{k}{n}\left(x-\frac{k}{n}\right)^2\mathrm{d}x\right|\\ &=\frac{Mn}{2}\left|\sum_{k=0}^n\int^0_{-\frac{1}{n}}x^2\mathrm{d}x\right|\\ &=\frac{Mn}{2}n\frac{1}{3n^3}=\frac{M}{6n}\rightarrow 0 \end{aligned}$

Thus,we have

$\begin{aligned} &\lim_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{k=0}^nf\left(\frac{k}{n}\right)-\int_0^1f(x)\mathrm{d}x\right)\\&=-\lim_{n\rightarrow\infty}n\sum_{k=0}^n\int_{\frac{k-1}{n}}^\frac{k}{n}f'\left(\frac{k}{n}\right)\left(x-\frac{k}{n}\right)\mathrm{d}x\\ &=-\lim_{n\rightarrow\infty}n\sum_{k=0}^nf'\left(\frac{k}{n}\right)\int_{-\frac{1}{n}}^0x\mathrm{d}x\\ &=\frac{1}{2}\lim_{n\rightarrow\infty}\sum_{k=0}^nf'\left(\frac{k}{n}\right)\frac{1}{n}\\ &=\frac{1}{2}\int_0^1f'(x)\mathrm{d}x\\ &=\frac{1}{2}[f(1)-f(0)] \end{aligned}$

RainField
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