problem concerning euler constant

Recently when I was browsing MSE, I came across a question about Euler constant.

The problem is:

How to show that $$ \lim_{n\to\infty}\left[\sum^n_{k=1}\frac{1}{k}-\ln(n)\right]=0.5772\ldots $$ No clue at all. Need help! Appreciated!

However, a certain part of the second answer confused me.

Abel’s partial summation technique:
Let $A(n) = \displaystyle \sum_{k=1}^n a(k)$. We then have

$$ \begin{align*} \sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\ & = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\ & = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n)) \end{align*} $$ (The above is nothing but the discrete version of integration by parts). $$ \sum_{n=1}^{N} a(n) f(n) = \int_{1^-}^{N^+} f(t) d(A(t)) = f(t) A(t) \rvert_{1^-}^{N^+} - \int_{1^-}^{N^+} A(t) f'(t) dt $$ (The second integral can be interpreted as a Riemann-Stieltjes integral.)

Here is the original post.

  The first part is clear. But for the second part, the equation $\displaystyle \sum_{n=1}^{N} a(n) f(n) = \displaystyle \int_{1^-}^{N^+} f(t) d(A(t))$ in particular, I was suprised to see such a peculiar expression of integral which I cann’t recognize at all and that a discrete summation could be turned into an integral so easily. So I asked Mr.Zhou Ruisong for help. He deemed that $\displaystyle \int_{1^-}^{N^+} f(t) d(A(t))$ stood for $\displaystyle \lim_{a\rightarrow 1^{-}} \lim_{b \rightarrow N^{+}} \int_{a^-}^{b^+} f(t) d(A(t))$ and that $A(n)=A(\lfloor n\rfloor)$ for real $n$, but with uncertainty. Along with his two sensible assumptions, I tried to wrote down a proof of the formular.

Proof Assume that $A:\mathbb{R}\rightarrow\mathbb{R}$ and $A(n)=A(\lfloor n\rfloor)$ for real $n$, $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous.

  Let’s consider the integer inteval $[1,N]$.

  Given $\epsilon>0$. Let a partition $P=\{x_0,x_1,\dots,x_n\}$ s.t. $U(f,P,A)-L(f,P,A)<\epsilon$,
$x_{i-1}<x_{i}$ $\forall i\in [1,n]$, $x_{0}=1-a$ where $a\in (0,1)$, $x_{n}=N+b$ where $b\in (0,1)$ and $\{x\vert x=k-\frac{1}{2},k\in \mathbb{N}^{*}\cap [1,N+1]\}\subset P$.

  Then,

$$ \begin{align*}\Bigg \vert \sum_{i=1}^{n} f(t_i)\big [ A(x_i)- A(x_{i-1})\big ] - \int_{1-a}^{N+b} fdA \Bigg \vert < \epsilon \qquad \forall t_{i}\in[x_{i-1},x_{i}].\end{align*} $$

  In particular, we let $t_{i}=\lfloor x_{i}\rfloor$ if $x_{i-1}<\lfloor x_{i}\rfloor < x_{i}$. Thus,

$$ \sum_{i=1}^{n} f(t_i)\big [ A(x_i)- A(x_{i-1})\big ]=\sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ]. $$

  Hence,

$$ \begin{align*} \Bigg \vert \sum_{i=1}^{n} f(t_i)\big [ A(x_i)- A(x_{i-1})\big ] - \int_{1-a}^{N+b} fdA \Bigg \vert= \Bigg \vert \sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ] - \int_{1-a}^{N+b} fdA \Bigg \vert < \epsilon.\end{align*} $$

  Since $\epsilon$ is arbitrary, we get

$$ \begin{align*} \sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ] =\int_{1-a}^{N+b} fdA.\end{align*} $$

  Finally, letting $N+b \rightarrow N^{+}$, $1-a \rightarrow 1^{-}$, we obtain

$$ \begin{align*} \sum_{i=1}^{N} f(i)\big [ A(i)- A(i-1)\big ] =\int_{1^{-}}^{N^{+}} fdA,\end{align*} $$

which is the desired result.

(Afterwards I discovered that $1^{-}$ denotes $1-\epsilon$, where $\epsilon>0$ is arbitrarily small and $N^{+}$ denotes $N+\epsilon$, where $\epsilon>0$ is arbitrarily small, according to the other answer of the author. The idea of the proof remains unchanged, though.)