TITLE: Proving $\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$
QUESTION [14 upvotes]: I've been trying to prove 
$$\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$$
I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find some pattern that could relate this to the harmonic series but this just seems surreal. I can't find any logical reasoning behind this identity. I don't understand Wikipedia's proof. Is there any proof that doesn't require the integral transform?

REPLY [2 votes]: The following variation is Example 3 in section 1.2  of John Riordan's classic Combinatorial Identities.

Consider for $n=1,2,\ldots$
  \begin{align*}
f_n&=\sum_{k=1}^n(-1)^{k+1}\binom{n}{k}\frac{1}{k}\\
&=\sum_{k=1}^n(-1)^{k+1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\frac{1}{k}\\
&=f_{n-1}-\frac{1}{n}\sum_{k=1}^n(-1)^k\binom{n}{k}\\
&=f_{n-1}-\frac{1}{n}\left[(1-1)^n-1\right]\\
&=f_{n-1}+\frac{1}{n}\\
&=H_n
\end{align*}
  since $f_1=1$.