Is there a more elementary way to arrive at: $\ln(m)=\lim_{n\to\infty}\sum_{k=n+1}^{mn}\frac{1}{k}$?

Question

Let $\psi$ denote the digamma function, and let $m,n\in\Bbb N_1$. For $m=1$, use the empty sum as $\ln(1)=0$ anyway!

The Gauss multiplication tells us that:

$$\sum_{i=1}^{m-1}\psi\left(n+\frac{i}{m}\right)=-m\ln m+\psi(mn+1)-\psi(n+1)$$

Recalling the harmonic relations of digamma, and allowing for loose limit notation, this becomes:

$$\begin{align}m\left(-\ln m+\sum_{k=1}^{mn}\frac{1}{k}-\sum_{k=1}^n\frac{1}{mk}\right)&=\sum_{k=1}^\infty\left(\frac{m-1}{k}-\sum_{i=1}^{m-1}\frac{m}{mk+mn-i}\right)\\-m\ln m+\sum_{k=1}^n\sum_{i=1}^{m-1}\frac{m}{mk-i}&=\sum_{k=1}^\infty\frac{m-1}{k}-\sum_{k=n+1}^\infty\sum_{i=1}^{m-1}\frac{m}{mk-i}\\-m\ln m&=\sum_{k=1}^\infty\left(\frac{m-1}{k}-m\sum_{i=1}^{m-1}\frac{1}{mk-i}\right) \end{align}$$

Introducing explicit limits, one finds:

$$\begin{align}-m\ln m&=\lim_{N\to\infty}\sum_{k=1}^N\left(\frac{m-1}{k}-m\sum_{i=1}^{m-1}\frac{1}{mk-i}\right)\\&=\lim_{N\to\infty}\sum_{k=1}^N\frac{m-1}{k}-m\left(\sum_{k=1}^{m\cdot N}\frac{1}{k}-\sum_{k=1}^N\frac{1}{mk}\right)\\&=\lim_{N\to\infty}-\sum_{k=N+1}^{m\cdot N}\frac{m}{k}\end{align}$$

That is:

$$\lim_{n\to\infty}\sum_{k=n+1}^{mn}\frac{1}{k}=\ln m$$

For all $m\in\Bbb N_1$. Amusingly, Desmos corroborates it for fractional $m$, but how exactly a fractional sum is computed there I am unsure!

I discovered this by accident whilst fiddling around with some other digamma calculations.

Is it in fact well known? If so, what are the more elementary ways of deriving it? I assume that if this is well known then more elementary ways exist.

Answer 1

$$\sum_{k=n+1}^{mn}\frac{1}{k}$$ is a Riemann Sum of the function $f(x)=\frac{1}{x}$ in the interval $[1,m]$, so $$\lim_{n\to\infty}\sum_{k=n+1}^{mn}\frac{1}{k} = \int_1^m\frac{1}{x}dx = \log(m)$$

Answer 2

If $n\gg 1$ then $\sum_{k=1}^{mn} 1/k \sim \ln mn+\gamma$, $\sum_{k=1}^{n}1/k\sim \ln n+\gamma$. Hence $$\sum_{k=n+1}^{mn} 1/k\sim \ln mn-\ln n =\ln m.$$ This can be easily converted to a $(\epsilon, N)$-proof. Take $n>N$ big enough so that $|Sum_1-\ln mn-\gamma|\lt \epsilon$, and $|Sum_2-\ln n-\gamma|\lt \epsilon$, hence $-\epsilon< Sum_1-\ln mn-\gamma <\epsilon$, $-\epsilon< Sum_2-\ln n-\gamma <\epsilon$, and subtract.

Answer 3

$$S:=\sum_{k=n+1}^{nm}\frac1k=\int_n^{nm}\frac{dt}{\lceil t\rceil}$$ so that

$$\int_n^{nm}\dfrac{dt}{t+1}< S< \int_n^{nm}\dfrac{dt}{t}.$$

The exact bounds are $\log\left(\dfrac{nm+1}{n+1}\right)$ and $\log(m)$.