If $f(x)=\sin x^2$ then what is $\lim_{n \to \infty} \left(\sum_{k=1}^{ n}\left(\frac{k}{n} \right) - n\int_0^1 f(x)dx \right) $

Question

Given that $f(x)=\sin x^2$ for $x \in \mathbb{R}$. Then how to find the following limit $$\displaystyle\lim_{n \to \infty} \left(\sum_{k=1}^{ n}f\left(\frac{k}{n} \right) - n\int_0^1 f(x)dx \right) $$

The expression is looking familiar,so I tried by multipling the denominator and the numerator with $\frac{1}{n}$ and have $$\displaystyle\lim_{n \to \infty} \frac{\frac{1}{n}\sum_{k=1}^{ n}f\left(\frac{k}{n} \right) - \int_0^1 f(x)dx }{\frac{1}{n}} $$

Now we know that $$\displaystyle\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{ n}f\left(\frac{k}{n} \right) =\int_0^1 f(x)dx $$ Hence it's a $\frac{0}{0}$ form .But how to proceed further from here? Also I would like to ask if there is any generalized method to solve this types of problems means if $S_n$ is the limit sum of an integral and $ S_n \rightarrow S$ as $n \rightarrow \infty$ then how to find $\displaystyle \lim_{n \to \infty}\frac{S_n-S}{\frac{1}{n}}$?

Do you happen to mean $$\sum_{k=1}^n f(k/n)$$ If you sum to infinity, then it is unclear to me why this should converge to the integral. — Severin Schraven, Jun 08 '25 at 09:36
Take {0,1/n,2/n,...n/n} the partition points of [0,1] and consider the Riemann sum with tag k/n for $x \in [k-1/n, k/n]$. — Mathson, Jun 08 '25 at 09:41
As Severin Schraven said, you ought to have $\sum_{k = 1}^{n}$ and not $\sum_{k = 1}^{\infty}$. A general method for this type of problem (if the integrand is sufficiently regular) is the Euler–Maclaurin formula. — Dermot Craddock, Jun 08 '25 at 10:29
A solution: https://math.stackexchange.com/questions/4870709 — Yutian, Jun 08 '25 at 10:53
@Mathson Indeed, you write it correctly, you take ${ 0/n, 1/n, \dots, n/n}$. These are $n$ intervals. The corresponding Riemann sum has $n$ terms, i.e. your sum cannot consist of infinitely many terms. — Severin Schraven, Jun 08 '25 at 12:25
Thanks for the kind informations.Euler–Maclaurin formula is helpful here . — Mathson, Jun 08 '25 at 18:57

Laughing Logarithm · Answer 1 · 2025-06-08T18:34:49.697

I supposed you meant $$ S_n := \lim_{n \to \infty} \left( \sum_{k=1}^{n} f\left( \frac{k}{n} \right) - n \int_0^1 f(x) \, dx \right) $$

Notice, \begin{align} S_n &= \sum_{k=1}^n f(k/n) - n \int_{(k-1)/n}^{k/n} f(x) dx \\ &= n \sum_{k=1}^n \int_{(k-1)/n}^{k/n} f(k/n) - f(x) dx \end{align}

As $f \in C^2(\mathbb{R})$, $$ f(x) = f(k/n) + f'(k/n) (x-k/n) + \frac{f''(\zeta_k)}{2} (x-k/n)^2$$ for some $ \zeta_k \in ( \frac{k-1}{n}, \frac{k}{n} ) $

Thus, \begin{align} S_n &= - \sum_{k=1}^n \int_{(k-1)/n}^{k/n} f'(k/n) (x-k/n) + \frac{f''(\zeta_k)}{2} (x-k/n)^2 dx \\ &= \sum_{k=1}^n \frac{f'(k/n)}{2n} - \frac{f''(\zeta_k)}{6n^2} \end{align} As $$ \left\|\sum_{k=1}^n \frac{f''(\zeta_k)}{6n^2} \right\| \leq \frac{1}{6 n} \sup_{[0,1]} |f''| \rightarrow 0 $$ Therefore, $$ S_n = \int_0^1 \frac{f'(x)} {2} dx + o(1) $$ In the case $ f(x) =sin(x^2 )$ $$ S_n \rightarrow \frac{sin(1)} {2} $$

I got it .But if you can explain the expansion of $f(x)=f(k/n)+....$ , it would be more helpful. Thank you. — Mathson, Jun 08 '25 at 18:55

If $f(x)=\sin x^2$ then what is $\lim_{n \to \infty} \left(\sum_{k=1}^{ n}\left(\frac{k}{n} \right) - n\int_0^1 f(x)dx \right) $

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