Relation between binomial coefficients and harmonic numbers

EDIT

I received notice that this question is a duplicate of A finite sum involving the binomial coefficients and the harmonic numbers

The answer is also included in https://en.wikipedia.org/wiki/Harmonic_number#Calculation

Original post

Let

$$t(\text{n},\text{k})\text{=}(-1)^{n-k} \binom{n}{k}$$

Prove that for n = 1, 2, 3, …

$$\sum _ {k = 1}^n \frac {t (n, k)} {k} = (-1)^{n + 1} \sum _ {k = 1}^n \frac {1} {k} = (-1)^{n + 1} H_n$$

Remark: It is interesting that the finite sum over the inverse integers is identical up to a factor $(-1)^{n+1}$ to the same sum over inverse integers each of which weighted with a binomial coefficient.

Solutions Collecting From Web of "Relation between binomial coefficients and harmonic numbers"

We start the proof with the interesting integral representation of HarmonicNumber (c.f. the solution of poweierstrass in How to prove the result of this integration?)

$$h(\text{n})\text{=}\int_0^1 \frac{1-(1-x)^n}{x} \, dx$$

Which can be proved easily by noticing that

$$h(1)=\int_0^1 \frac{1-(1-x)}{x} \, dx=1$$

and showing that the characteristic recursion relation for HarmonicNumber

$$h(n+1)=h(n)+\frac{1}{n+1}$$

holds.

In fact,

$$h(n+1)=\int_0^1 \frac{1-(1-x) (1-x)^n}{x} \, dx=\int_0^1 \frac{1-1 (1-x)^n}{x} \, dx+\int_0^1 \frac{x (1-x)^n}{x} \, dx\\=h(n)+\frac{1}{n+1}$$

qed.

Now in the integral $h(n)$ we expand the term $(1-x)^n$ binomically, interchange integration and summation, and find

$$h(n)=\int_0^1 \frac{1-\sum _{k=0}^n (-1)^k x^k \binom{n}{k}}{x} \, dx=\sum _{k=1}^n (-1)^{k+1} \binom{n}{k} \int_0^1 x^{k-1} \, dx\\=\sum _{k=1}^n \frac{(-1)^{k+1} \binom{n}{k}}{k}=(-1)^{n+1} \sum _{k=1}^n \frac{t(n,k)}{k}$$

Which completes the proof.
QED.

Remark: I have included this result as a formula to https://oeis.org/A130595.