<s>[INST] Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$ Calculate  $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$, I do not know hot get rid of that $k$, for me it is similar like  $\binom{n}{k}=\frac{k}{n} \binom{n-1}{k-1}$, do you have some idea?
 [/INST] This problem and its type appear  at MSE regularly. Suppose we seek to
compute
$$S_n = \sum_{k=1}^n {n\choose k} \frac{(-1)^{k+1}}{k}.$$
With this in mind we introduce the function
$$f(z) = n! (-1)^{n+1} \frac{1}{z^2} \prod_{q=1}^n \frac{1}{z-q}.$$
We then obtain for $1\le k\le n$
$$\mathrm{Res}_{z=k} f(z) =
(-1)^{n+1} \frac{n!}{k^2} \prod_{q=1}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = (-1)^{n+1} \frac{n!}{k}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= {n\choose k} \frac{(-1)^{k+1}}{k}.$$
This means that
$$S_n = \sum_{k=1}^n \mathrm{Res}_{z=k} f(z)$$
and since residues sum to zero we have
$$S_n + \mathrm{Res}_{z=0} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$
We can compute  the residue at infinity by inspection  (it is zero) or
more formally through
$$\mathrm{Res}_{z=\infty}
n! (-1)^{n+1} \frac{1}{z^2} \prod_{q=1}^n \frac{1}{z-q}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0} \frac{1}{z^2}
z^2 \prod_{q=1}^n \frac{1}{1/z-q}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0}
\prod_{q=1}^n \frac{z}{1-qz}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0} z^n
\prod_{q=1}^n \frac{1}{1-qz} = 0.$$
We get for the residue at $z=0$ that
$$\mathrm{Res}_{z=0} f(z) =
n! (-1)^{n+1}
\left. \left(\prod_{q=1}^n \frac{1}{z-q}\right)'\right|_{z=0}
\\ = - n! (-1)^{n+1} \left.
\left(\prod_{q=1}^n \frac{1}{z-q}\right)
\sum_{q=1}^n \frac{1}{z-q} \right|_{z=0}
\\ = n! (-1)^n \frac{(-1)^{n}}{n!} \left(-H_{n}\right)
= -H_n.$$
We thus have $S_n - H_n = 0$ or
$$\bbox[5px,border:2px solid #00A000]{
S_n = H_n = \sum_{k=1}^n \frac{1}{k}.}$$
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