Show $\lim_{n \to \infty}\prod\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right) = e^{\int_{0}^{1}f(x)dx}$ where $f(x)$ is continuous

Question

I need to show: $$\lim_{n \to \infty}\prod\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right) = e^{\int_{0}^{1}f(x)dx}$$

My attempt:

$\lim_{n \to \infty}[\sum_{k=1}^{n}\log(1+\frac{1}{n}f(\frac{k}{n}))-\log(e^{\int_{0}^{1}f(x)dx})= 0 $

$\lim_{n \to \infty}[\sum_{k=1}^{n}\log(1+\frac{1}{n}f(\frac{k}{n}))-\int_{0}^{1}f(x)dx]= 0 $

$\lim_{n \to \infty}[\sum_{k=1}^{n}\log(1+\frac{1}{n}f(\frac{k}{n}))-\log(e^{\sum_{k=1}^{n}\frac{1}{n}f(\frac{k}{n})})]= 0 $

$\lim_{n \to \infty}[\sum_{k=1}^{n}\log(\frac{1+\frac{1}{n}f(\frac{k}{n})}{e^{\frac{1}{n}f(\frac{k}{n})dx}})] = \lim_{n \to \infty}\log((\frac{1+\frac{1}{n}f(\frac{1}{n})}{e^{\frac{1}{n}f(\frac{1}{n})}}))+\cdots+\lim_{n \to \infty}\log((\frac{1+\frac{k}{n}f(\frac{k}{n})}{e^{\frac{1}{n}f(\frac{k}{n})}}))+ \cdots + \lim_{n \to \infty}\log((\frac{1+\frac{1}{n}f(\frac{n}{n})}{e^{\frac{1}{n}f(\frac{n}{n})}})) \cdots (1)$

(Since $\log(x)$ is a continuous function) $\lim_{n \to \infty}log((\frac{1+\frac{1}{n}f(\frac{k}{n})}{e^{\frac{1}{n}f(\frac{k}{n})}})) = log(\lim_{n \to \infty}(\frac{1+\frac{k}{n}f(\frac{k}{n})}{e^{\frac{1}{n}f(\frac{k}{n})}}))$

We know that $\lim_{n \to \infty}\frac{1}{n}f(\frac{k}{n})\to 0$

Hence it is of the form $\lim_{x \to 0}\frac{e^{x}-1}{x} = 1$

Then $\lim_{n \to \infty}\sum_{k=1}^{n}\log(1+\frac{1}{n}f(\frac{k}{n}))= \lim_{n \to \infty} \sum_{k=1}^{n}\log(1) = 0$

Is my approach correct?

Answer 1

I'm not sure what you are doing from the 3. line onwards (the sum is in the exponent, not term-wise), but starting with the second line and assuming finiteness of $f(x)$ on $[0,1]$ and $n$ being large you can estimate the following $$\sum_{k=1}^n \log\left(1+\frac{1}{n}\,f\left(\frac{k}{n}\right)\right) = \sum_{k=1}^n \left[ \frac{1}{n}\,f\left(\frac{k}{n}\right) + O\left(1/n^2\right) \right] \\ = \frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right) + O(1/n) \, .$$

Answer 2

You can’t go from $\lim_{n\to\infty}\sum_{k=1}^n$ to $\sum_{k=1}^n\lim_{n\to\infty},$ because the number of terms in the sum depends on $n.$ The sum doesn’t even make sense outside the limit, because there is no value of $n$ there.

In particular, all Riemann integrals would be zero if you could do this. If $f(x)$ is bounded on $[0,1],$ then $$\int_0^1 f(x)\,dx =\lim_{n\to\infty}\sum_{k=1}^{n}\frac1nf(k/n)$$ But since $f$ is bounded, $\frac1nf(k/n)\to 0$ for any $k.$

So your argument would give us zero for all Riemann integrals.

As a commenter noted, use that $\log(1+x)=x+O(x^2),$ for $|x|<1.$

One way get a limit of zero for $$\lim_{n\to\infty}\sum_{k=1}^n a_{n,k}\tag1$$ is to show the terms satisfy $$\lim_{n\to\infty}n\max_{k\in\{1,2,\dots,n\}} |a_{n,k}|=0$$ for all $k.$

In your case: $$a_{n,k}=\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)-\frac{1}{n}f\left(\frac{k}{n}\right)$$

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Another approach is to use the dominated convergence theorem.

If $\lim_{n\to\infty}a_{n,k}= 0$ for all $k,$ and $b_k$ is a sequence such that $|a_{n,k}|\leq b_k$ for all $k,n,$ and $\sum b_k<+\infty,$ then (1) converges to $0$.