<s>[INST] Proving $\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$ 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? 
 [/INST] Suppose we seek to verify that
$$\sum_{k=1}^n {n\choose k} (-1)^{k+1} \frac{1}{k} = H_n.$$
Now observe that
$$\frac{1}{k} =
\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \log\frac{1}{1-z} \; dz$$
where we take $\epsilon < 1$ so that the logarithmic term is analytic.
We get for the sum
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z} \log\frac{1}{1-z} 
\sum_{k=1}^n {n\choose k} (-1)^{k+1} \frac{1}{z^k}
\; dz$$
The term for $k=0$ yields
$$-\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z} \log\frac{1}{1-z} = 0$$
and we may lower $k$ to zero, getting for the sum
$$-\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z} \log\frac{1}{1-z} 
\sum_{k=0}^n {n\choose k} (-1)^{k} \frac{1}{z^k}
\; dz
\\ = -\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z} 
\left(1-\frac{1}{z}\right)^n
\log\frac{1}{1-z} 
\; dz
\\ = -\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(z-1)^n  \log\frac{1}{1-z}  \; dz.$$
Now put $$\frac{z}{z-1} = w
\quad\text{so that}\quad
z = -\frac{w}{1-w} 
\\ \text{and}\quad
dz = - \frac{1}{(1-w)^2} dw
\quad\text{and}\quad
\frac{1}{1-z} = 1-w.$$
to get
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(z-1)^{n+1} \frac{1}{1-z}  \log\frac{1}{1-z}  \; dz
\\ = - \frac{1}{2\pi i} 
\int_{|w|=\gamma} \frac{1}{w^{n+1}}
(1-w) \log(1-w) \frac{1}{(1-w)^2} \; dw
\\ = \frac{1}{2\pi i} 
\int_{|w|=\gamma} \frac{1}{w^{n+1}}
\frac{1}{1-w} \log\frac{1}{1-w}  \; dw.$$
This evaluates to $$H_n$$ by inspection since
$$\log\frac{1}{1-z} = \sum_{q\ge 1} \frac{z^q}{q}$$
as seen earlier.
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