How Can I Compute $\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \dfrac{1}{k}$?

Question

Proof

$$ \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dots + \dfrac{1}{n} $$

This might be a silly question asked by someone else but I cannot find it right now.

I tried to verify this formula in Mathematica, and I am sure that this statement is correct.

The first approach I tried is to use generating functions and the binomial theorem, where I got

$$ \dfrac{\partial}{\partial x}(-\sum _{k=1}^n \dfrac{x^k}{k} \binom{n}{k}) = -\sum _{k=1}^n x^{k-1} \binom{n}{k} = \dfrac{1-(x+1)^n}{x} $$

But I am having difficulties integrating the RHS: even Mathematica gives a function that includes "HypergeometricPFQ" which is not elementary.

The second approach I tried is to use induction. But I find it really difficult to use the inductive hypothesis.

Is there a better way to solve this question? Or where can I find any resources?

But

Does this answer your question? Prove by induction: $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ - found using an Approach0 search. Note there also quite a few other duplicates (with all of the following questions actually ... — John Omielan, Aug 29 '23 at 03:34
Yes. I am sorry for posting duplicate questions. Thank you for helping! — Linyuan Gong, Aug 29 '23 at 03:36
(cont.) being closed as duplicates of other questions) here, e.g., Prove by induction $\sum _{i=1}^n\left(-1\right)^{i+1}:\binom{n}{i}:\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$, Proof of a sum with binomial coefficients $\sum_{k=1}^n (-1)^{k+1}{\binom nk}\frac{1}{k} = 1 + \frac{1}{2} + \ldots +\frac{1}{n}$, Proof for any positive integer n the following equality hold, Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$, ... — John Omielan, Aug 29 '23 at 03:39
(cont.) Prove that $\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k} = 1+\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$, etc. Finally, you're welcome for the help, and also welcome to Math SE. — John Omielan, Aug 29 '23 at 03:39

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