Euler-Mascheroni constant: understanding why $\lim_{m\rightarrow \infty} \sum_{n=1}^{m} (\ln (1 + \frac{1}{n})-\frac{1}{n+1})= 1 - \gamma$

Question

I am trying to understand why the Euler-Mascheroni constant $\displaystyle \gamma = \lim_{n \rightarrow \infty} \left ( \sum_{k=1}^n \frac{1}{k} - \ln n \right )$ is equal to $1 - \displaystyle \int_{1}^{\infty} \frac{t-[t]}{t^2}\; \mathrm{d}t$ where $[t]$ is the floor function.

There has been an answered question here already:
Integral form for the euler-mascheroni gamma constant using floor function

However, there is one line I do not understand and will appreciate help on explaining why the following equality is true: $$ \lim_{m\rightarrow \infty} \sum_{n=1}^{m} (\ln (1 + \frac{1}{n})-\frac{1}{n+1})= 1 - \gamma $$

Answer 1

Note that

$$\ln \left(1+\frac{1}{n}\right) = \ln \frac{n+1}{n} = \ln (n+1) - \ln n.$$

Hence

$$\begin{align} \sum_{n=1}^m \left(\ln \left(1+\frac{1}{n}\right) - \frac{1}{n+1}\right) &= \sum_{n=1}^m (\ln (n+1) - \ln n) - \sum_{n=1}^m \frac{1}{n+1}\\ &= \underbrace{\ln (m+1) - \sum_{k=1}^{m+1} \frac{1}{k}}_{\to -\gamma} + 1. \end{align}$$

Answer 2

However, there is one line I do not understand and will appreciate help on explaining why the following equality is true: $$\lim_{m\to\infty} \sum_{n=1}^m \left( \ln \left(1 + \tfrac{1}{n}\right) - \frac{1}{n+1}\right) = 1-\gamma$$

Observe that $\ln \left(1 + \frac{1}{n}\right) = \ln (n+1) - \ln n \implies \displaystyle\sum_{n=1}^m \ln \left(1 + \frac{1}{n}\right) = \ln (m+1)$ (it telescopes).

Therefore $$\sum_{n=1}^m \left(\ln \left(1 + \tfrac{1}{n}\right) - \frac{1}{n+1}\right) = \ln (m+1) - \sum_{n=1}^{m+1} \frac{1}{n} + 1$$

and $$\ln (m+1) - \sum_{k=1}^{m+1} \frac{1}{n} \to -\gamma$$ as $m\to\infty$ by definition.