**Case 2**

An analogous procedure can be exploited to handle the second:

$$
\hat{\mu}^r \eta\_0 = \frac{{}^{1}h\_r}{r} \tag{51}
$$

which allows for the derivation of the following identity

$$\sum\_{n=0}^{\infty} \frac{t^n}{n!} \,\_2h\_{n+m} = \varepsilon^t \sum\_{r=0}^m \binom{m}{r} \partial\_{t-2}^r \phi(t) \,\_r \tag{52}$$

with

$$\begin{cases} \,\_2\phi(t) = 1 - e^{-\hat{\mu}\cdot t} \eta\_0, \\ \,\_2\partial\_{t}\phi(t) = \begin{cases} -\sum\_{s=0}^{\infty} \frac{(-1)^{s+r}}{s!(s+r)} \,\_1h\_{s+r} \, \,\_r^s & r > 0, \\\ -\sum\_{r=1}^{\infty} \frac{(-t)^r \,\_1h\_r}{r!r} \,\_r & r = 0. \end{cases} \end{cases} \tag{53}$$

**Remark 1.** *Let us now use the obvious identity (which holds for all operators):*

$$1^n = \left[\left(1 - \hat{\mu}\right) + \hat{\mu}\right]^n \eta\_0. \tag{54}$$

*Expanding the Newton binomial, we find*

$$\begin{split} [(1-\hat{\mu})+\hat{\mu}]^n \eta\_0 &= \sum\_{r=0}^n \binom{n}{r}\_r \hat{\mu}^{n-r} \sum\_{s=0}^r \binom{r}{s} (-1)^s \hat{\mu}^s \eta\_0 = \\ &= \left( \sum\_{r=0}^{n-1} \binom{n}{r} \sum\_{s=1}^r \binom{r}{s} (-1)^s \hat{\mu}^{n+s-r} + \sum\_{r=0}^{n-1} \binom{n}{r} \hat{\mu}^{n-r} + \sum\_{s=1}^n \binom{n}{s} (-1)^s \hat{\mu}^s + 1 \right) \eta\_0. \end{split} \tag{55}$$

**Remark 2.** *Choosing, for example, the first operator* (46) *and the property <sup>κ</sup>*<sup>ˆ</sup>*rκ*<sup>ˆ</sup>*<sup>s</sup>ψ*0 = *<sup>κ</sup>*<sup>ˆ</sup>*s*+*<sup>r</sup>ψ*0 = 1 *s* + *r, we can finally elaborate* (54) *to get (see [24]):*

$$\sum\_{r=1}^{n} \binom{n}{r} \frac{1}{r} = -\sum\_{k=1}^{n} \binom{n}{k} \sum\_{r=1}^{k} \binom{k}{r} (-1)^{r} \frac{1}{n+r-k} \tag{56}$$

*while when repeating the procedure with the realization* (51)*, we end up with*

$$\sum\_{r=1}^{n} \binom{n}{r} \frac{1}{r} \frac{h\_r}{r} = -\sum\_{k=1}^{n} \binom{n}{k} \sum\_{r=1}^{k} (-1)^r \binom{k}{r} \frac{1}{n+r-k}.\tag{57}$$

According to Reference [6] (see Equation (2)), it might also beconvenient to use the following umbral definition for the inverse of an integer.

**Definition 3.** *We introduce the umbral operator*

$$
\hat{a}^n \gamma\_0 := \frac{1}{n+1}, \qquad \forall n \in \mathbb{N}.\tag{58}
$$

**Proposition 3.** ∀*n* ∈ N *(see Equation* 1 *in [6])*

$$\frac{1}{n+1} = \sum\_{s=0}^{n} \binom{n}{s} (-1)^{s} \frac{1}{s+1}.\tag{59}$$

**Proof.** It can be proved by induction that from (For *n* = 2 we find *<sup>a</sup>*ˆ2*γ*0 = -1 − 2*a*<sup>ˆ</sup> + *a*ˆ2 *γ*0, according to the prescription in Equation (58), we ge<sup>t</sup>

$$\frac{1}{2+1} = 1 - 2\frac{1}{1+1} + \frac{1}{2+1};$$

for *n* = 3, *<sup>a</sup>*ˆ3*γ*0 = -1 − 3*a*<sup>ˆ</sup> + 3*a*ˆ2 − *a*ˆ3 *γ*0, we ge<sup>t</sup>

$$\frac{1}{4} = 1 - 3\frac{1}{1+1} + 3\frac{1}{2+1} - \frac{1}{3+1}.$$

$$
\hbar^n \gamma\_0 = (1 - \hbar)^n \gamma\_0. \tag{60}
$$

Furthermore, by using the obvious relation

$$\hbar^{\mathrm{n}}\gamma\_{0} = \left[1 - (1 - \hbar)\right]^{\mathrm{n}}\gamma\_{0} = \sum\_{s=0}^{\mathrm{n}} \binom{\mathrm{n}}{s} (-1)^{s} \left(1 - \hbar\right)^{s},\tag{61}$$

we end up with Equation (59).

> We consider now the square of harmonic numbers, [1*hn*]2.

**Definition 4.** *We introduce the umbral operator*

$$\prescript{}{1}{h}\_{\begin{pmatrix} 2 \end{pmatrix}}^n \prescript{}{2}{\xi}\_0 := \prescript{}{h}{h}^n \prescript{}{\xi}\_0 = \begin{bmatrix} \_1h\_n \end{bmatrix}^2, \quad \forall n \in \mathbb{N}. \tag{62}$$

**Definition 5.** *The umbral operator* (62) *can be exploited to define the formal series:*

$$\tilde{\zeta}\_{h^{\tilde{\mathbb{K}}}(2)}e(t) := \overline{e}(t) = 1 + \sum\_{n=1}^{\infty} \frac{t^n}{n!} \tilde{h}^n \overline{\zeta}\_0 = e^{\int\_{\mathbb{T}} t} \overline{\zeta}\_{0\prime} \tag{63}$$

*specifying the associated generating function.*

The derivation of Equation (63), according to the previously foreseen method, reduces to the solution of a first-order differential Equation, as it has been shown in the following example.

**Example 4.** *By noting that*

$$h^{n+1}\mathcal{J}\_0 = \left[h\_n\right]^2 + 2\frac{h\_n}{n+1} + \frac{1}{(n+1)^2} \tag{64}$$

*and that*

$$\begin{aligned} \, \, \partial\_{\underline{h}} \, \, \tilde{c}(t) = \tilde{h} \, \, \epsilon^{\underline{h} \cdot \underline{t}} \, \, \tilde{\xi}\_{0} \,. \end{aligned} \tag{65}$$

*we easily find*

$$\partial\_t \vec{e}(t) = \vec{e}(t) - 1 + \frac{2}{t} \sum\_{n=0}^{\infty} \frac{t^{n+1}}{(n+1)!} \left( {}\_1 h\_{n+1} - \frac{1}{n+1} \right) + \sum\_{n=0}^{\infty} \frac{t^n}{(n+1)\left(n+1\right)!} \tag{66}$$

˜

*or*

$$\partial\_t \left. \overline{e}(t) \right| = \overline{e}(t) + \frac{2}{t} (\,\_1\overline{e}(t) - t - 1) - \frac{1}{t} \int\_0^t \frac{e^s - s - 1}{s} \, ds \tag{67}$$

*which can be solved by the use of the same technique as before, made only slightly more complicated by the non-homogeneous term. The Equation can be straightforwardly solved, and the relevant solution, for e*˜(0) = 1*, reads*

$$
\vec{e}(t) = \mathfrak{e}^t \left( 1 + \chi(t) \right),
\tag{68}
$$

*where*

$$\begin{aligned} \chi(t) &= \int\_0^t e^{-s} \beta(s) \, ds, \\ \beta(t) &= \frac{2}{t} (\, \_0\text{e}(t) - t - 1) - f\_2(t). \end{aligned} \tag{69}$$

**Remark 3.** *In the introductory section we mentioned the link between harmonic and Bernoulli numbers, and they have been paradigmatic examples of applications of umbral methods. Regarding the finite sum of power of integers, we find, for example (according to Chapter 6 in Reference [4]),*

$$\sum\_{m=0}^{n} r^{m} = \,\_{(-m)}h\_{\text{ul}} = \frac{1}{m+1} \left[ \left( \hat{b} + n + 1 \right)^{m+1} - \hat{b}^{m+1} \right] \theta\_{0\prime} \tag{70}$$

*where*

$$
\hat{b}^r \theta\_0 = B\_r \tag{71}
$$

*is the umbral operator which provides Bernoulli numbers Br.*

**Proposition 4.** *The use of Equation* (70) *yields the generating function:*

$$\sum\_{m=0}^{\infty} \frac{(-m)^{h\_m} z^{m+1}}{m!} = \frac{z}{c^z - 1} (e^{(n+1)z} - 1), \qquad \forall z \in \mathbb{R} : |z| < 2\pi. \tag{72}$$

**Proof.** ∀*z* ∈ R :| *z* |< 2*π*

$$\sum\_{m=0}^{\infty} \frac{(-m)^{h\_n} z^{m+1}}{m!} = \sum\_{m=0}^{\infty} \frac{\left[ (\hat{b} + n + 1)^{m+1} - \hat{b}^{m+1} \right] z^{m+1}}{(m+1)!} \theta\_0 = \left( e^{(\hat{b} + n + 1)z} - e^{\hat{b}z} \right) \theta\_0 = 0$$

$$= \left( e^{(n+1)z} - 1 \right) e^{\hat{b}z} \theta\_0 = \left( e^{(n+1)z} - 1 \right) \sum\_{r=0}^{\infty} \frac{z^r}{r!} \, \hat{b}^r \theta\_0 =$$

$$= \left( e^{(n+1)z} - 1 \right) e^{\hat{b}z} \theta\_0 = \left( e^{(n+1)z} - 1 \right) \sum\_{r=0}^{\infty} \frac{z^r}{r!} \, \hat{B}\_r$$

which, by taking into account that the generating function of Bernoulli numbers is

$$\sum\_{r=0}^{\infty} \frac{z^r}{r!} B\_r = \frac{z}{e^z - 1} \tag{73}$$

yields the statement in Equation (72).

In this paper, we have provided a hint of the possible interplay between the theory of harmonic numbers and operational methods (be they of umbral ordinary nature). Further progress will be presented in future investigations.

**Author Contributions:** Conceptualization, G.D.; methodology, G.D., S.L.; data curation: S.L.; validation, G.D., S.L., E.S., H.M.S.; formal analysis, G.D., S.L., E.S., H.M.S.; writing—riginal draft preparation: G.D.; writing—review and editing: S.L.

**Funding:** The work of S.L. was supported by a *Enea-Research Center Individual Fellowship*

**Acknowledgments:** The Authors recognizes the kind help of the Referees who pushed for a better organized presentation and for including in the study the Bernoulli numbers.

**Conflicts of Interest:** The authors declare no conflict of interest.
