Expressing a function in terms of an infinite series involving its derivative

Question

Let $G\left(x\right) = \int{g\left(x\right) \,dx}$, therefore:

$$\int_{a}^{b}{g\left(x\right) \,dx} = G\left(b\right) - G\left(a\right) = \left(b - a\right)\lim_{N \to \infty}{\frac{1}{N}\sum_{n = 1}^{N}{g\left(a + \frac{n}{N}\left(b - a\right)\right)}}$$

Let $g\left(x\right) = f'\left(x\right)$:

$$f\left(b\right) - f\left(a\right) = \left(b - a\right)\lim_{N \to \infty}{\frac{1}{N}\sum_{n = 1}^{N}{f'\left(a + \frac{n}{N}\left(b - a\right)\right)}}$$

Let $b = x$:

$$f\left(x\right) = f\left(a\right) + \left(x - a\right)\lim_{N \to \infty}{\frac{1}{N}\sum_{n = 1}^{N}{f'\left(a + \frac{n}{N}\left(x - a\right)\right)}}$$

I believe this would be valid for $f\left(x\right)$ differentiable on the interval $(a, x]$.

(1) Let $a = x_{0} : f\left(a\right) = 0$:

$$f\left(x\right) = \left(x - x_{0}\right)\lim_{N \to \infty}{\frac{1}{N}\sum_{n = 1}^{N}{f'\left(x_{0} + \frac{n}{N}\left(x - x_{0}\right)\right)}}$$

(2) Let $a = 0$:

$$f\left(x\right) = f\left(0\right) + x\lim_{N \to \infty}{\frac{1}{N}\sum_{n = 1}^{N}{f'\left(\frac{nx}{N}\right)}}$$

Using (2), let $f\left(x\right) = \ln\left(1 + x\right)$:

$$\ln\left(1 + x\right) = x\lim_{N \to \infty}{\sum_{n = 1}^{N}{\frac{1}{N + nx}}}$$

which yields that, for $x = 1$:

$$\ln\left(2\right) = \lim_{N \to \infty}{\sum_{n = 1}^{N}{\frac{1}{N + n}}}$$

Although this formula requires a lot of terms to reach even a few significant figures of accuracy, I think it is beautiful to think that such a constant can be represented in such a consise way.

My question is, is this derivation correct? And, if so, would any results from this be at all useful, given that the rate of convergence is so slow?

ADDITION

$$\int{\frac{f\left(x\right) - f\left(a\right)}{x - a} \,dx} = \int{\frac{f\left(x\right)}{x - a} \,dx} - f\left(a\right)\ln{|x - a |} = C + \lim_{N \to \infty}{\sum_{n = 1}^{N}{\frac{1}{n}f\left(a + \frac{n}{N}\left(x - a\right)\right)}}$$

Let $a = 0$:

$$\int{\frac{f\left(x\right)}{x} \,dx} - f\left(0\right)\ln{|x|} = C + \lim_{N \to \infty}{\sum_{n = 1}^{N}{\frac{1}{n}f\left(\frac{nx}{N}\right)}}$$

Let $x = Nt$, $N \to \infty$:

$$\int{\frac{f\left(Nt\right)}{t} \,dt} - f\left(0\right)\ln{|Nt|} = C + \sum_{n = 1}^{N}{\frac{1}{n}f\left(nt\right)}$$

Let $f\left(x\right) = 0$, therefore $C = 0$:

$$\int{\frac{f\left(Nt\right)}{t} \,dt} - f\left(0\right)\ln{|Nt|} = \sum_{n = 1}^{N}{\frac{1}{n}f\left(nt\right)}$$

Answer 1

The derivation is indeed correct.

Note that we can verify independently that $\log(2)=\lim_{N\to \infty}\sum_{n=1}^N\frac{1}{n+N}$ without appealing to Riemann sums.

To do so, we write

$$\begin{align} \sum_{n=1}^N \frac{1}{N+n}&=\sum_{n=N+1}^{2N}\frac1n\\\\ &=\sum_{n=1}^{2N}\frac1n-\sum_{n=1}^{N}\frac1n\\\\ &=\sum_{n=1}^N\left(\frac{1}{2n-1}+\frac{1}{2n}\right)-\sum_{n=1}^{N}\frac1n\\\\ &=\sum_{n=1}^N\left(\frac{1}{2n-1}-\frac{1}{2n}\right) \tag 1\\\\ &=\sum_{n=1}^N \frac{(-1)^{n-1}}{n}\\\\ &=\sum_{n=1}^N (-1)^{n-1}\int_0^1 x^{n-1}\,dx\\\\ &=\int_0^1 \sum_{n=0}^{N-1} (-x)^{n}\,dx\\\\ &=\int_0^1 \frac{1-(-x)^N}{1+x}\,dx \tag 2 \end{align}$$

Note that upon taking the limit as $N\to \infty$ in $(2)$, we have

$$\begin{align} \lim_{N\to \infty}\int_0^1 \frac{1-(-x)^N}{1+x}\,dx &=\int_0^1 \lim_{N\to \infty}\left(\frac{1-(-x)^N}{1+x}\right)\,dx\\\\ & =\int_0^1\frac{1}{1+x}\,dx\\\\ &=\log(2) \end{align}$$

as was to be shown! The interchange of the limit and the integral can be justified by applying, for example, the Dominated Convergence Theorem.

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To answer the question in the OP regarding speed of convergence, note that the form in $(1)$, upon taking the limit as $N\to \infty$, provides an absolutely convergent series that has terms that behave as $\frac{1}{2n^2}$ for large $n$.