Is this formula for the harmonic numbers true?

Question

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[%]

Answer 1

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.