Limit of differences of truncated series and integrals give Euler-gamma, zeta and logs. Why? In the MSE-question in a comment to an naswer Michael Hardy brought up the following well known limit- expression for the Euler-gamma
$$  \lim_{n \to \infty} \left(\sum_{k=1}^n \frac 1k\right) - \left(\int_{t=1}^n \frac 1t dt\right) = \gamma \tag 1$$
I've tried some variations, and heuristically I found for small integer $m \gt 1$
$$  \lim_{n \to \infty} (\sum_{k=1}^n \frac 1{k^m}) - (\int_{t=1}^n \frac 1{t^m} dt) = \zeta(m) - \frac 1{m-1} \tag 2$$
With more generalization to real $m$ it seems by Pari/GP that eq (1) can be seen as a limit for $m \to 1$ and the Euler-$\gamma$ can be seen as the result for the Stieltjes-power-series representation for $\zeta(1+x)$ whith the $\frac 1{1-(1+x)}$-term removed and then evaluated at $x=0$      

Q1: Is there any intuitive explanation for this (or, for instance, a graphical demonstration)?                

Another generalization gave heuristically also more funny hypotheses:
$$ \tag 3$$
$$ \small \begin{eqnarray}  
\lim_{n \to \infty} (\sum_{k=2}^n \frac 1{k(k-1)}) &-& (\int_{t=2}^n \frac 1{t(t-1)} dt) &=& \frac 1{1!} \cdot(\frac 11 - 1\cdot \log(2)) \\
\lim_{n \to \infty} (\sum_{k=3}^n \frac 1{k(k-1)(k-2)}) &-& (\int_{t=3}^n \frac 1{t(t-1)(t-2)} dt) &=& \frac 1{2!} \cdot(\frac 12 - 2\cdot \log(2) + 1\cdot \log(3) ) \\
\lim_{n \to \infty} (\sum_{k=4}^n \frac 1{k...(k-3)}) &-& (\int_{t=4}^n \frac 1{t...(t-3)} dt) &=& \frac 1{3!} \cdot(\frac 13 - 3\cdot \log(2) + 3\cdot \log(3)- 1\cdot \log(4) ) \\
 \end{eqnarray} $$
where the coefficients in the rhs are the binomial-coefficients and I think the scheme is obvious enough for continuation ad libitum.
Again it might be possible to express this with more limits: we could possibly write, for instance the rhs in the third row as
$$ \lim_{h\to 0} \frac 1{3!} \cdot(- \small \binom{3}{-1+h}  \cdot \log(0+h) +1 \cdot \log(1) - 3\cdot \log(2) + 3\cdot \log(3)- 1\cdot \log(4) ) \tag 4$$

Q2: Is that (3) true and how to prove (if is it not too complicated...)? And is (4) somehow meaningful?

 A: For Q1. the proof just relies on summation by parts.
For Q2., you can evaluate
$$S_k = \sum_{n=1}^{+\infty}\frac{1}{n(n+1)\ldots(n+k)} = \frac{1}{k!}\sum_{n=1}^{+\infty}\frac{1}{n\binom{n+k}{k}}$$
by exploiting partial fractions decomposition and the residue theorem, or just the wonderful telescoping trick $\frac{1}{n(n+k)}=\frac{1}{k}\left(\frac{1}{n}-\frac{1}{n+k}\right)$, giving:
$$\begin{eqnarray*}S_k &=& \frac{1}{k}\left(\sum_{n=1}^{+\infty}\frac{1}{n(n+1)\ldots(n+k-1)}-\sum_{n=1}^{+\infty}\frac{1}{(n+1)(n+2)(n+k)}\right)\\ &=&\frac{1}{k}\cdot\frac{1}{1\cdot 2\cdot\ldots\cdot k}=\frac{1}{k\cdot k!}.\end{eqnarray*}$$
The same telescoping technique applies to the integral:
$$I_k = \int_{1}^{+\infty}\frac{dt}{t(t+1)\ldots(t+k)}=\frac{1}{k}\int_{0}^{1}\frac{dt}{(t+1)\ldots(t+k)}$$
and now the RHS can be evaluated through partial fraction decomposition, since:
$$\frac{1}{(t+1)\ldots(t+m)}=\frac{1}{(m-1)!}\sum_{j=0}^{m-1}\frac{(-1)^j\binom{m-1}{j}}{t+j+1}.$$
We have $\int_{0}^{1}\frac{dt}{t+h}=\log(h+1)-\log(h)=\log\left(1+\frac{1}{h}\right)$, hence:
$$\begin{eqnarray*}I_k &=& \frac{1}{k(k-1)!}\sum_{j=0}^{k-1}(-1)^j\binom{k-1}{j}\left(\log(j+2)-\log(j+1)\right)\end{eqnarray*}$$
just gives your $(3)$ after rearranging terms.
