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**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

- Harmonic number inequality
- Proof by induction: $\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$
- Harmonic Numbers series I
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- Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$
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**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.

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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.

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