<s>[INST] Is this formula for the harmonic numbers true? Is this formula for the harmonic numbers true?
$$H_n = \lim_{s\to 0} \, \int \frac{(s+1)^{(-n-1)}+s-1}{s} \, ds$$
Mathematica:
Clear[n, s]
Monitor[Table[
  Limit[Integrate[((s + 1)^(-n - 1) + s - 1)/s, s], s -> 0], {n, 1, 
   12}], n]
Differences[%]

 [/INST] I claim that
$$\frac{1}{(s+1)^n s} = \frac{1}{s} - \sum_{j=1}^{n} \frac{1}{(s+1)^j}.$$
Let us prove this by induction. If $n=1$ then one may check that 
$$\frac{1}{s (s+1)} = \frac{1}{s} - \frac{1}{s+1}$$
as desired. Now assume the statement is true for $n=k$. Using the inductive hypothesis we get
$$\frac{1}{(s+1)^{k+1} s} = \frac{\frac{1}{(s+1)^k s}}{s+1} = \frac{1}{s (s+1)} - \sum_{j=2}^{k+1} \frac{1}{(s+1)^j}.$$ 
Performing the same decomposition again gives
$$\frac{1}{(s+1)^{k+1} s} = \frac{1}{s} - \sum_{j=1}^{k+1} \frac{1}{(s+1)^j}$$
as desired.
Using the claim, we have
$$\frac{1}{(s+1)^{n+1} s} + 1 - \frac{1}{s} = 1 - \sum_{j=1}^{n+1} \frac{1}{(s+1)^j}$$
Now for the original problem we have
$$\int \frac{1}{(s+1)^n s} + 1 - \frac{1}{s} ds = s - \ln(s+1) + \sum_{j=2}^{n+1} \frac{1}{(j-1) (s+1)^{j-1}} + C$$
so the limit is
$$\sum_{j=2}^{n+1} \frac{1}{j-1} + C = H_n + C$$
Note that the second $s$ in the numerator is in some sense superfluous, since the result of the limit is the same with or without it.
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