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.
 A: 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.
